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Hello, in the last class we have learnt boundary
value problems with derivative boundary
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conditions.
.
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Now, let us learn higher order boundary value
problems, higher order BVPs, so let us
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start with an example first consider this
equation, so y dashed of 0 is 1 and say y
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of 1 is
minus 1, say h is 1 by 3. Now, with usual
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discretization one must obtain for y 3 what
should be the corresponding approximation,
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so we have done it in the first lecture.
So, accordingly y 3 is approximated as minus
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2, y i is 3 x i square, now if we run this
equation for i 1, y 3 is 2 y 2 plus 2 y 0
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minus y 1 minus 4 h cube y i 6 h cube x i
square
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and at i 2, 2 y 1 and this will be y 0 minus
4 is y 1 y 2 six x 2 square. So, what was
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our
line, our line was x 0 is 0 and x 1, x 2,
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x 3 is 1. So, we have the following fictitious
values, which are corresponding to the nodes
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outside the actual grid. Now, let us look
at
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the boundary conditions, so the boundary conditions
this by prime.
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..
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Let us discretize the boundary conditions,
so we start as follows so these are fictitious
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then y dashed of 1 is minus 1 this is y 4
y 2 by 2 h minus 1 this implies then we have
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y 0
equals to 1 and then y dashed of 0 equals
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this implies. So, we have these two fictitious
values. Now, if you look at
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suppose we call this a this y 0 is explicitly
given, so this gets
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determined then we have y 1, y 2, y 3 there,
so y 0 gets determined so y 1, y 2, y 3. So,
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three unknowns then this fictitious value
is determined and this is determined, however
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A B C is not a tri diagonal that means there
is some issue with respect to the number of
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unknowns and number of equations, well following
earlier strategy.
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.
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.Suppose, we write we write at a i equals
to 0, i equals to 3 we get y minus 2 and y
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5
which are fictitious. So, it means that one
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cannot adopt the earlier strategy like you
run
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the equation earlier in the sense which we
have done with the derivative boundary
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conditions. So, one cannot adopt the similar
strategy because in a sense of recurs manner
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you get number of fictitious values if you
run further i minus 1, i minus 4 you get more
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fictitious values, so this strategy does not
help.
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.
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Let us look in to more general case, so this
is more general case. So, we consider third
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order linear equation given by y 3 plus, so
when I said general we have all the terms,
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so
that makes life little complex, so 3 2 1 and
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y 0. So, all the terms are there and since
it is
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third order equation we need a 3 boundary
conditions, so y of a is alpha gamma. So,
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the
grid points
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x n, x n plus 1 equals to b, now what are
the approximations, so for first
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approximation
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second
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and for the third.
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..
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Now, with these approximations, so one gives
say a x i, we get this and boundary
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conditions y 0 is alpha is beta. As I mentioned
four and five is not tri diagonal system,
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when we write four at i is 0, N we get the
fictitious values, so these are the fictitious
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values. However, one can obtain three equations,
we already have three equations from
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the boundary conditions five still, and we
are left with and left with one more that
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is one
more equation. So, how do we supply, so here
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we bank on the order of the equation see
the order of the equation is higher order
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and we know how to reduce it to a system.
.
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.So, we use these let y dash is p then we
get the equation
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and correspondingly y of a is
alpha and p of a is beta and p of b is gamma
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so that is how our system gets reduced.
Now, what we do is we discretize the new,
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so what kind of discretization we propose,
so
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discretization of new as follows
this we discretize, so backward there, so
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this is average
or some trapezoidal integration. So, this
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is for p and for other equation usual central,
so
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what did we achieve, we will see what we have
achieved.
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We have a taking on the higher order equation
we have converted into a system, so this
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serves as one more equation this serves as
the one more equation that we require. So,
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let
us see how to proceed, so from six
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we have
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and seven, sorry a i, y i, this is a i y i
plus b i
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p i minus 1 plus c i p i plus d i look at
that this contain p i minus 1 p i plus 1 and
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y i.
.
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So, this is written in this from, where a
i is c of x i b i is
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and
this can be verified very
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easily. Now, we have to somehow adjust the
number of equations and the number of
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unknowns, so how do we do it. Let us proceed
step by step, so let us write down
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six and
seven for i equal to 1, sorry eight this six
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and seven means or eight and nine. So, from
here when we write for i equals to 1 y 0 plus
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h by 2, p 0 y 1, p 1 equals to 0 and from
here a 1 y 1, b 1 p 0, c 1 p 1, d 1 p 2, so
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these can be written asâ€¦
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..
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How did we do that, look we have y 0 alpha
and p 0 is beta, so these are known we have
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pulled it to the right hand side. Then from
the other one
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then we have to be careful, so
what I have done these two, eight and nine
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we have written for i equal to 1. Now, these
two that means these two we need to run from
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two onwards. So, let us look at it if we run
for two y 1 and p 1, y 2, p 2, so you have
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form this we have a vector what is a vector
y i
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p i kind of vector it is see, look at it y
i minus 1 p i minus 1 y i p i.
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So, at least the index is matching, but whereas
here there is slight deviation, thus y i p
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i
minus i p i plus 1, so let us look at it,
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how we can place this? So, this can be written
as
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you introduce some notation, let us introduce
notation x n is x i, x i y i, p i. So, if
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we
introduce this, so in general any x n is y
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n, so there is a coefficient.
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..
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.
If you look at it suppose look at this equation
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y i minus 1, p i minus 1 if you like to
reproduce this coefficient of this must be
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1 there and coefficient of this must be h
by 2
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there. Similarly, for y i and p i so that
means from here, if we plot for y i minus
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1 there is
no coefficient. Therefore, we expect 0 there
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where as for pi minus 1 we have b i, so what
did we achieve these two equations.
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So, two terms we have produced and these two
terms, yes we have produced so that
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means these two equations can be put in a
matrix form. So, this further can be 1 h by
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2 b
i and as per our notation this can be x i
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where x i is x i minus 1 y i minus 1 p i minus
1.
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So, what we are going to do, we have to follow
very carefully these two equations for 2
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to n, we are trying to put it in the matrix
from and with a vector notation, so if we
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do that
we get.
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..
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This can be written as
A i plus B i plus C i equals D i and what
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is our notation X i then
we should be able to write down the matrix
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A, so again I go back
x i minus 1 which
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means y i minus 1 p i minus 1. So, we need
the coefficient matrix a i for this, so what
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should be A i the coefficients coming from
first equation the corresponding coefficients
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from second equation, so which we already
have written down look at it.
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So, here one comes and from here h by 2 from
there here no y i minus 1 before 0 and we
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have p i minus 1 we have b i, so this must
be our a i, so what was our a i 1 h by 2 0
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b i.
Let us look at next we need b i x i and what
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is x i y i p i look at that so here we have
minus 1 minus 1 for y i h by 2 for p i.
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..
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So, which means
this minus 1 there h by 2 there this is multiplied
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y i p i, look at this and
what is multiplying here a i is multiplying
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y i and c i is multiplying. So, we are getting
the corresponding coefficients so the two
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equations in the matrix form.
.
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So, B i must be minus 1 then C i
X i plus 1, so X i plus 1 look at that, so
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here we do not
have any y i plus 1 p i plus 1. So, the first
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entry must be 0 whereas, here we have p i
plus
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1 therefore, so I am not explaining like this
because it is it is a straightforward. Now,
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there is no y i plus 1 p i plus 1, so their
first entries must be 0 the matrix, but whereas
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.here there is no y i plus 1. So, the first
entry of the second must be 0 and we have
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D i
there, so correspondingly C i must be 0 0
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0 d i because it is multiplying x i plus 1.
Then, d i must be column matrix we have exhausted
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all the terms here and here and then
the third one there is nothing on the right
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hand side where as here we have third one.
There is nothing on the right hand side where
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as we have this term while this be this
notation wise this is matrix this is the unknown
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function. So, there are equations and this
from i 2 3 and minus 1 and these are how many
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n minus 2 equations, but we had written
something for one, so for one we have written,
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so this also can be putted in matrix form
probably.
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.
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So, let us put it
i equal to 1 case can be Written as B 1 X
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1, C i X 2, so we have not have
to give some number 12, 10 and 11 i equals
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to 1, 10 and 11. They can be putted like this
how, look at that here we have y 1 p 1 which
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is nothing but exactly or vector x 1, so from
here y 1 p 1 is a 1 c 1, so our matrix b is
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more or less structured. So, b 1 minus 1 h
by 2 a
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1 c 1 this is small c coefficient then the
matrix c 1 matrix c 1.
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.We do not have the next because c is multiplying
x i plus 1, so when i is 1 x 2, but we do
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not have any two terms there c is multiplying
x i plus 1. So, when i is 1 x 2, but we do
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not have any two terms there, so we get 0
0 and 0 d 1 and d 1 is minus alpha minus h
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by
2 beta minus b 1 beta plus h square d of x
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1. So, this is 13, now we have covered i 1
i 2
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00:36:02,380 --> 00:36:03,810
to n minus 1.
.
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00:36:03,810 --> 00:36:59,450
Then we are left with i n, now i n case gives
us however, so from where we would get
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this. Of course, from here the matrix, so
this can be putted as a n x n minus 1, where
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see
x n minus 1 that is y n minus 1 and p n minus
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1. So, we have the terms very much one
and here we do not have y n minus 1, so p
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n minus 1 has the term then b n, look the
remaining left out we know p n what was p
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n gamma it is a boundary condition because
y dashed of 1 which is p n. So, this term
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00:38:31,880 --> 00:38:39,880
is gone to the right hand side d n, so we
can
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expect minus 1 there and there is no as such
p n left, where as we here we have y n and
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p
n plus 1.
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00:38:58,470 --> 00:39:10,950
So, we had look at carefully this is n minus
1, so y n minus 1 p n minus 1 is done then
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x
n, so we have to write x n. So, x n should
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contain y n and p n, so p n is known, so we
shift to the other side and even here we know
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00:39:29,280 --> 00:40:02,350
it. So, this will be a n, d n with a definition
x n equals y n, p n plus 1 because this is
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known to us, so we push it to this side.
Therefore, unknowns y n and p n plus 1 that
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is a small adjustment, so you can make a
note of this, because the symmetry is lost
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and then we are in position.
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.So, d n must be p n is known, so it has gone
minus gamma by 2 h and here we know this
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00:40:40,690 --> 00:41:12,200
is gamma, so minus c n gamma we have. So,
let us compile n minus 2 equations and then
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this is a another equation, so n minus 1 and
this is another one equation. So, total n
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00:41:31,300 --> 00:41:47,190
equations, but in what sense they are n minus
2 plus 1 plus 1 and each is two equations
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and equations in
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00:42:19,380 --> 00:42:30,240
how many unknowns 2 n unknowns because these
are two and these are
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00:42:30,240 --> 00:42:38,410
two times n minus 1. So, what is the achievement
by putting in this matrix form by
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00:42:38,410 --> 00:42:47,600
converting into system the number of unknowns
and the numbers of equations have been
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00:42:47,600 --> 00:43:05,430
matched. Now, eliminating the unknowns we
get, let us see carefully look at this, so
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00:43:05,430 --> 00:43:17,870
d is
for 1 to n minus 1, so then this is for 1.
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00:43:17,870 --> 00:43:18,870
.
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00:43:18,870 --> 00:43:34,260
So, if we write a matrix this is all matrices
right, so suppose B b 1 and C 1 there this
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00:43:34,260 --> 00:43:52,530
must multiply. Let us say in a bigger matrix
these two must multiply x 1, x 2 then 2 so
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00:43:52,530 --> 00:44:04,850
for 2 to n minus 1 it is like this, so let
us start the index to 8 must multiply x 1
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00:44:04,850 --> 00:44:12,880
so that
means we must expect 8 there then b 2 must
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00:44:12,880 --> 00:44:30,040
multiply x 2 b 2 there then c 2.
So, this continues each of them is column
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00:44:30,040 --> 00:44:45,440
vector for example, what is x 1 and what are
they these are matrices in turn, so this corresponds
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00:44:45,440 --> 00:44:55,100
to and equals to 1 and i equals to 1
and i 2 to n minus 1 and the last what we
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00:44:55,100 --> 00:45:01,820
obtained the last one i equals to n. So, this
will
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00:45:01,820 --> 00:45:12,540
also come at the bottom and this entire thing
which one I am talking about this is you see
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00:45:12,540 --> 00:45:16,453
it is a block, but then again it is a tri
diagonal system.
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00:45:16,453 --> 00:45:17,453
..
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00:45:17,453 --> 00:45:29,520
So, it is a block tri diagonal system, so
we get as follows A X equals to D some global
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00:45:29,520 --> 00:45:39,400
where the structure of A follows from the
earlier matrices, these are null matrices
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then A
2, B 2 and C 2 like this.
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.
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00:46:09,190 --> 00:46:45,110
What is our X, it is better I write X looks
like this see x 1 is x 2. So, this is our
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00:46:45,110 --> 00:46:57,260
X and
so
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00:46:57,260 --> 00:47:18,530
this is
in fact I should write here this is a block
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00:47:18,530 --> 00:47:27,340
tri diagonal system, so with a reference to
BVP, what did we do.
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00:47:27,340 --> 00:47:28,340
..
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00:47:28,340 --> 00:47:34,240
If you discretize the equations definitely
we are short of one equation and this supplied
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by really writing the higher order equation
in terms of a system. So, this system will
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00:47:41,360 --> 00:47:44,720
help
us to supply the additional one more equation
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00:47:44,720 --> 00:47:50,240
and since this is a system we have to solve
simultaneously, so we obtain block tri diagonal
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00:47:50,240 --> 00:47:56,450
matrix, so let us try with an example.
.
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00:47:56,450 --> 00:48:33,460
So, this simplest, so for calcification purpose
I have taken h equals to 1 say 0, 1, 2, 3,
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so
then with usual discretization, so discretization
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00:49:03,460 --> 00:49:20,300
of this 0 discretization of this
because
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00:49:20,300 --> 00:49:28,339
two the denominator goes there.
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00:49:28,339 --> 00:49:29,339
..
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00:49:29,339 --> 00:50:25,060
Now, we write one at i equals to 1, so this
then i equals to 2, so however here and from
205
00:50:25,060 --> 00:50:53,150
boundary conditions we have, so this gives
us. So, for example in this y 0 is given and
206
00:50:53,150 --> 00:50:57,500
y
minus 1 can be substituted y for can be substituted.
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00:50:57,500 --> 00:51:26,990
So, this whole thing leaves us with y
1, y 2, y 3 as unknowns, now at i equals to
208
00:51:26,990 --> 00:52:20,760
0, y 2 y 1 y 0 at i equals to 3, so what
happened? So, these are new fictitious, so
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00:52:20,760 --> 00:52:30,590
two equations, so because these are all
redundant. Now, this reduces to two equations,
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00:52:30,590 --> 00:52:43,680
but y 1, y 2, y 3 and five unknowns.
.
211
00:52:43,680 --> 00:53:02,840
.We have to really depend on y dash equals
to p, so then p double plus y is x. So,
212
00:53:02,840 --> 00:53:44,320
correspondingly this one and p of 0 is 0,
so with the discretization of this we get
213
00:53:44,320 --> 00:53:46,880
so that
means discretization of this discretization
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00:53:46,880 --> 00:54:15,640
of this so and then we have let us have some
number. So, i equals to 1 a gives us 2 y 1
215
00:54:15,640 --> 00:54:38,210
minus p 1 is 2 because I have simplified you
can do it then B gives us right, so this is
216
00:54:38,210 --> 00:54:43,750
we have done for it a general case. Now, for
A
217
00:54:43,750 --> 00:55:04,580
this A and B for i 2 and 3, so this can be
written as, in fact i 2 only one we have because
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00:55:04,580 --> 00:55:16,859
our grid was 0, 1, 2 and 3, so this is x 0,
x 1, x 2 and x 3, so this can be written as
219
00:55:16,859 --> 00:55:24,740
D i.
.
220
00:55:24,740 --> 00:55:55,960
So, we if we try to do that we arrive at
221
00:55:55,960 --> 00:57:06,460
so this is 1 set and another set. Ultimately,
we get
222
00:57:06,460 --> 00:57:14,650
another matrix here and this one, so this
will be the tri diagonal system which we have
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00:57:14,650 --> 00:57:18,720
to
solve. So, this gives the structure for the
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00:57:18,720 --> 00:57:23,520
higher order boundary value problems, so like
initial value problems higher order one reduce
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00:57:23,520 --> 00:57:26,670
it to system and can solve it. Similarly,
we
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00:57:26,670 --> 00:57:31,890
have been reducing to system to support one
more equation, but then we end up with
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00:57:31,890 --> 00:57:39,869
block tri diagonal system which one can solve
it. So, there are other methods for solving
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00:57:39,869 --> 00:57:44,940
non linear boundary value problems, so we
look at them in the coming lectures, until
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00:57:44,940 --> 00:57:45,940
then, bye.
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00:57:45,940 --> 00:57:45,940
.