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