Good afternoon everyone, so we will be starting
this class with the view of whatever we have
finished in the last class. And in the last
class, what we did is that we covered, we
looked into, examined, some of the standard
equations in Cartesian coordinate, where the
equation which is valid for cylindrical coordinate.
The equation which is valid for the spherical
polar coordinate and Euler's equation.
We have looked into different you know solution
of different types of equations, under different
set of boundary conditions, may be a Dirichlet
boundary condition, may be a Neumann boundary
condition or Robin mixed boundary condition.
So, we have seen how depending on the square
boundary conditions, the solution of such
problems are changed and the eigenvalues and
eigenfunctions become different for different
boundary conditions.
And except the Cartesian coordinate, we have
also looked into the Basel equation and Legendre
function, Legendre equation. And we have seen
how the eigenvalues and eigenfunction will
be appearing in the form of Basel function
or Legendre polynomial. And we have also looked
into different properties this Basel functions
or Legendre polynomial, they will obey...
And also we have seen into the solution of
Euler's equation, not only that, we looked
into the, we have developed the theorem for
to get the adjoint operator, given an operator
L.
Now, what will be doing in this class, will
be formulating, the we will carry forward
the development of the eigenvalue problem
and will be defining a standard eigenvalue
problem or a Sturm-Liouville problem. Once
we define a standard eigenvalue problem or
Sturm-Liouville problem, will be looking into
some of the theorems and axioms these eigenvalues
and eigenfunctions will obey. And these properties
we will be utilizing frequently when we will
be solving the equations using a partial differential
equations, using the separation of variable
type of solution.
Once, we complete the relevant theorems and
axioms of eigenvalues and eigenfunctions and
then will get into the actual solution of
partial differential equations. Let us start
the formulation of standard eigenvalue problem
or Sturm-Liouville problem.
For that we consider the parameter or operator
L L u is equal to a 0 x plus a 0 x d square
u d x square plus a 1 x d u d x plus a 2 x
u. Suppose we consider a function like this,
where the operator L is nothing but a 0 x
d square d x square plus a 1 x d d x plus
a 2 x. Now, on separation, after doing separation
of variables, we will be getting an equation
something like this, a 0 x d square u d x
square plus a 1 as a function of x d u d x
plus a 2 as a function of x u plus lambda
a 3 u is equal to 0.
Just consider this equation and will be doing
the separation of, actual separation of variables
later on. And for that time being, just take
this equation as granted, it is basically
in the form of L u is equal to minus lambda
a 3 u, so this is a generalized form of the
equation. In a particular equation, a 1 may
be 0 or a 2 may be 0, but we are going to
find out the different for different values
of lambda, how this equation will be transformed
into a standard eigenvalue problem.
Now, we assume a Dirichlet boundary condition
homogenous. If you remember in the last class,
we have looked into several types of boundary
conditions, but all these boundary conditions
are homogenous in nature. So, the generality
of the solution does not change, the steps
remains same and the formulation remains constant
if we change a homogenous Dirichlet boundary
condition to homogenous robin mixed or homogenous
Neumann boundary condition. Let us consider
that, we used a Dirichlet homogenous 
boundary condition and these are at x is equal
to 0, u is equal to 0, at x is equal to 1,
u is equal to 0.
So, let us proceed with this, now the form
of the equation is L u plus lambda a 3 u is
equal to 0. So, you will be getting L u is
equal to minus lambda a 3 u and if you look
into the similarity of the eigenvalue problem
in discrete domain, so you just remember,
recall the eigenvalue problem 
in discrete domain that was a x is equal to
lambda x.
So, this equation can be rewritten as this
form, so equation 1 is recast as d d x of
p of x d u d x plus q of x u plus lambda r
as a as a function of x times u is equal to
0, let say this is equation number 2.
Now, these two equations will be identical,
because we define p q r such that p of x is
e to the power integral a 1 x divided by a
0 x d x, q of x is nothing but a 2 x divided
by a 0 times p and r of x is nothing but a
3 x divided by a 0 x times p. Now, we can
substitute p q and r in this equation and
will be getting back this equation. So, let
us try to do that and prove that the form
of L u is equal to minus lambda a 3 u and
this form, that is equation 1 and 2 are identical
in nature.
So, therefore, we start with this, so p of
x is defined as e to the power integral a
1 x divided by a 0 x d x. If you take logarithm
on both side, this becomes l n p, is nothing
but a 1 x, integral a 1 x a 0 x d x. So, if
we differentiate both sides, you will be getting
with respect to x, you will be getting 1 over
p d p d x, is nothing but a 1 divided by a
0 and you will be getting d p d x is a 1 divided
by a 0 times p.
Now, if we put these values into the governing
equation d d x of p of x d u d x plus q of
x u plus lambda r of x u is equal to 0. Now,
what we are going to do, we open up open this
equation up, that means we differentiate this
part, so this becomes p of x d square u d
x square plus d p d x d u d x plus q u plus
lambda r u is equal to 0. So, therefore, we
write the equation as p of x, so this becomes,
substitute the different values of p and q
and r.
So, p of x d square u d x square plus d p
d x, we have already found that found out
that d p d x in the earlier one is a 1 over
a 0 times p of x times d u d x plus q, q we
have defined as a 2 by a 0 times p plus lambda
r u and r we have defined as a 3 by a 0 times
p is equal to 0. Now, if I have seen that
in all the terms we have a constant quantity,
that is p of x, we divide both sides by p
of x and multiply both side by a 0. So, what
you will be getting is a 0 as a function of
x d square u d x square plus a 1 function
of x d u d x plus a 2 function of x u plus,
there is one u there, so u plus lambda a 3
u is equal to 0.
So, if we now compare this equation with the
earlier one, that you had earlier as L u is
equal to minus lambda a 3 u, so this becomes
a 0 d square u d x square, write the operator
L, so this becomes a 1 d u d x plus a 2 u
plus, we take lambda a 3 on the other side,
so lambda a 3 u is equal to 0. So, these two
equations are identical, therefore this equation
can be equivalently written as d d x of p
of x d u d x plus q of x u plus lambda r,
which is, which can be in general function
of x times u. So, my operator, we can write
it as d d x of p d d x plus q, so this becomes
my operator and we can look into the adjoint
operator to this particular operator.
So, the operator becomes now L is equal to
d d x of p d d x plus q, so we have already
seen that earlier. We have in the last class,
we have seen that if L of v is a 0 v double
prime, that means d square v d d x square
plus a 1 v prime plus a 2 v, then the adjoint
operator L star v equal to a 0 v double prime
plus 2 a 0 prime minus a 1 times v prime plus
a 0 double prime minus a 1 prime plus a 2
times v. That means if my L is equal to a
0 d square d x square plus a 1 d d x plus
a 2, then my L star is a 0 d square d x square
plus 2 a 0 prime minus a 1 d d x plus a 0
double prime minus a 1 prime plus a 2.
So, therefore, we can compare these equation
with the earlier one, the operator was written
as d d x of p d d x plus q, therefore p d
square x, d square d x square plus d p d x
d d x plus q. So, therefore, we can compare
this equation with this equation and if we
can compare that a 0 becomes, if you compare,
let say these, right, let us write it sees
3, this is 4, so you can compare 3 and 4 and
can get different values.
So, therefore, we can by comparing we can
write p is a 0 is nothing but p of x, your
a 1 is nothing but d p d x and a 2 is nothing
but q of x. So, if we look into the L star,
the L star is a 0 d square d x square plus
2 a 0 prime minus a 1 d d x plus a 0 double
prime minus a 1 prime plus a 2. So, a 0 becomes
p, so this becomes p d square d x square plus
2 a 0 prime, that means 2 d p d x minus a
1, a 1 is d p d x d d x plus a 0 double prime
is d square p d x square minus a 1 prime is
d square p d x square plus a 2 is q. So, this
two will be cancelling out, so you will be
getting p d square d x square plus, it will
be one d p d x of d d x plus q, so this will
be d d x multiplied by p d d x plus q.
Now, this will be if you look into the earlier
slide, will be seeing that L is equal to L
star in this particular problem. So, we are
talking about a self-adjoint operator, this
particular operator in general is known as
this, in this particular operator in generally
is known as the Sturm-Liouville operator.
So, let us write down the Sturm-Liouville
equation and we have to if we have to prove
that if the Sturm-Liouville equation is a
self-adjoint equation, we have already proved
that L is equal to L star, but we have to
prove that b is equal b star as well, the
boundary operator should also match.
So, let us define now Sturm-Liouville operator
or Sturm-Liouville equation. This Sturm-Liouville
equation 
is also known as standard eigenvalue problem.
So, Sturm-Liouville equation is defined as
S L equation, it is defined as a 0 d square
u d x square plus a 1 d u d x plus a 2 u is
equal to minus lambda a 3 u.
So, subject to the boundary conditions, will
use to general boundary condition, that is
at x is equal to a, alpha 1 d u d x plus alpha
2 u is equal to 0 and at x is equal to b,
we have beta 1 d u d x plus beta 2 u is equal
to 0. So, we consider two most generalized
boundary condition, that if alpha 1 and beta
1 are 0, then both of this boundaries are,
they will be boiling down to Dirichlet boundary
condition. If alpha 2 equal to 0 and beta
2 is equal to 0, will be getting a Neumann
boundary condition, if both of them are non-zero,
then will be getting a robin mixed boundary
condition.
So, we can look into the operator, the operator
in this problem is nothing but a 0 d square
d x square plus a 1 d d x plus a 2, both a
1 a 0 a 1 a 2 they are function of x and this
equation can be written in this form. Already
we have seen that, that equation as Sturm-Liouville
equation, can be written in this form d d
x of p of x d u d x plus q of x u is equal
to minus lambda r of x times u, so L u is
equal to minus lambda r u.
So, this is a standard eigenvalue problem
in continuous function, continuous domain.
We have already seen that L is self-adjoint
in the just earlier to this, we have already
seen that L is equal to L star, so the operator
is self-adjoint. Now, in the earlier class,
we have seen this, if the system become self-adjoint,
then the operator has to self-adjoint, as
well as the boundary condition has to be self-adjoint.
That means boundary operator must be is equal
to, b is equal to b star, so for that you
have to, what you have to do, we have to examine
the bilinear concomitant J u v.
So, we have to look into the bilinear concomitant
term. So, if you look into that will be getting,
we will be writing J u v is equal to v a 0
u prime minus v a 0 prime of u plus a 1 v
u evaluated from on the boundaries a to b.
And this prime denotes differentiation with
respect to x, so if you just open this up,
this becomes v a u, v a 0 u prime minus v
prime a 0 u minus v a 0 prime u plus a 1 v
u from a to b.
We have already proved earlier that a 0 prime
is d p d x and this is p prime and a 1 is
nothing but d p d x is equal to p prime and
that is equal to a 0 prime. So, a 1 is equal
to p prime, a 0 is equal to p and a 0 prime
is equal to a 1. So, if we since a 0 prime
is equal to a 1, then last two terms of this
bilinear concomitant they will vanish, so
they will just out they will just out, they
will be cancelled each other.
So, what is the form of bilinear concomitant?
This will be v a 0 u prime minus v prime a
0 u evaluated between a to b. So, I can take
a 0 common, so this becomes v u prime minus
v prime u evaluated between a to b.
Now, will be simplifying it and check whether
to make this bilinear concomitant to 0, what
will be the conditions on v we have imposed.
Now, let us put, let us evaluate bilinear
concomitant term. So, this becomes a 0 v u
prime at evaluated at b minus v prime evaluated
at b u evaluated at b minus v evaluated at
a u prime evaluated at a minus minus plus
v prime evaluated at a u evaluated at a.
Now, we can recall the boundary conditions
on u, at x is equal to a, we have, if we recall
the boundary conditions on u, it will be alpha
1 u prime plus alpha 2 u is equal to 0 and
at x is equal to b, we have alpha beta 1 u
prime plus beta 2 u is equal to 0. So, therefore,
we substitute u prime a and u prime b from
this equation. So, v of b and what is u prime
b? u prime b is nothing but minus beta 2 by
beta 1 u at b minus v prime b and u b minus
v of a and what is u prime a? u prime a we
substitute as minus alpha 2 by alpha 1 times
u at a plus v prime a and u a remain as they
are.
So, just simplify this equation, this becomes
minus beta 2 by beta 1 u of b v of b, then
we have minus u of b v prime of b, then minus
minus plus alpha 2 by alpha 1 u of a v of
a plus v prime a and u prime a multiplication
on that. So, we have a 0, we take minus u
of b as common, minus and also beta 1, to
be divided by beta 1, minus u by b divided
by beta 1 we take as common.
So, what will be getting is that b 2 v of
b plus beta 1 v prime of b and from this two,
we combine this two, will be getting plus
u of a is common divided by alpha 1, so you
will be getting alpha 2 u of a plus alpha
1 alpha 2 v of a plus alpha 1 v prime of a.
Now, we do not have any idea about what is
the value of u evaluated at a and u on the
boundary x is equal to b. So, therefore, in
order to make this bilinear concomitant to
be vanished, the term in the second bracket,
they should be put equal to 0, individually
each of them.
So, if you do that, then what will be getting
is that beta 1 d v d x plus beta 2 v is equal
to 0 at x is equal to b and from the other
one, alpha 1 d v d x plus alpha 2 v is equal
to 0 at x is equal to a. So, therefore, these
two boundary conditions on v they emerge out
from the by putting J bilinear concomitant
equal to 0. And if we remember this is b star,
this is the boundary operator of the adjoint
problem, and if we remember the boundary conditions
b on the original problem, that was at x is
equal a alpha 1 d u d x plus alpha 2 u is
equal to 0 and at x is equal to b alpha 1
d v d x plus alpha 2 u is equal to 0.
So, beta 1 d u d x plus beta 2 u is equal
to 0 at x is equal to b. Therefore, both B
and B star are identical and we had L is equal
L star already proved earlier, so, therefore
the we proved that Sturm-Liouville problem
is 
problem or standard eigenvalue problem, has
is self-adjoint problem.
So, you will be having L is equal to L star
and B is equal to B star and therefore this
proves that Sturm-Liouville problem or standard
eigenvalue problem is a self-adjoint problem.
And standard eigenvalue operator or Sturm-Liouville
operator is a self-adjoint operator, so it
does not matter what kind of boundary conditions
it have, we have consider the most general
robin mixed boundary condition from which
the Dirichlet and Neumann conditions are specially
derivable under special conditions.
So, in the most general conditions of boundary
condition and the governing equation, the
Sturm-Liouville problem is a self-adjoint
problem.
So, therefore, if we have a so if we can now
identify what are characteristics of a Sturm-Liouville
problem, there are two characteristics of
Sturm-Liouville problem. The first one is
that, if the form of the equation must be
in this form L u is equal to minus lambda
r u or may be a function of x in general.
So, that is the form of the equation, governing
equations and the boundary conditions are
homogeneous. If these two conditions are satisfied,
these two characteristics are satisfied, then
we can have a Sturm-Liouville problem.
Next, we will be looking into some of the
theorems of eigenvalues and eigenfunctions.
The first theorem goes like this, there is
a countable infinity 
of 
eigenvalues lambda that means lambda must
be lying between minus infinity to plus infinity,
such that lambda n tends to infinity if n
tends to infinity. So, we should this statement
is equivalent that there are infinite 
number of eigenvalues exists in n dimensional
space. And in case of function, continues
function this n dimension becomes too large,
it becomes very large, it tends to infinity.
So, in case of continuous functions, there
are countable, but infinite number of eigenvalues
present.
Next, we look into theorem 2. This theorem
says that if lambda m and lambda n are two
distinct eigenvalues 
corresponding to 
eigenfunctions 
y m and y n, then the eigenfunctions y m and
y n are orthogonal functions with respect
to 
weight function r. So, these are for this
lambda m and lambda n are the eigenvalues
corresponding to Sturm-Liouville equation.
So, let us write down the Sturm-Liouville
equation, this is L y is equal to minus lambda
r, which is in general function of x and multiplied
by y.
Now, subject to the boundary operator B is
equal to 0, for x lying between small a and
small b, so let us assume lambda m and lambda
n are distinct eigenvalues there and the corresponding
eigenfunctions are y m and y n, are corresponding
eigenfunctions. So, this y m and y n must
satisfy this equation, when y becomes y m
then lambda becomes lambda m, when y become
y n lambda becomes lambda n.
So, therefore, we should write these two equations
as L y m is equal to minus lambda m r y m
and L y n is equal to minus lambda n r y n.
So, this is equation number 1, this is equation
number 2, we take inner product of equation
number 1 with respect to n y n and see what
we get. We get y n, inner product of y n and
L y m is equal to minus lambda m r y m, inner
product between these two.
Now, lambda m being a constant, it will be
coming out of the equilibrium sign, inner
product sign with the minus, so this becomes
inner product of r y m and y n. So, if you
remember in case of continuous function y
m y n d x integration, y m y n d x is nothing
but the inner product of y m and y n and this
is identical to inner product of y n and y
m.
So, what will be getting out of this, so you
will be again take the, so this is number
1 equation number 3. Then we take inner product
of equation 2 with respect to y m, so if you
do that you will be getting inner product
of y m L y n should be is equal to minus lambda
n inner product of r y n coma y m, this is
equation number 4.
Now, what we do, we subtract equation number
4 from 3 and see what we get. So, if we subtract
equation 4 from equation 3, will be getting
inner product of y n, L y m minus inner product
of y m and L y n. And this will be minus lambda
m inner product of r y m y n, can be written
as this integral r y m y n d x minus minus
plus lambda n r y m y n d x.
So, if we now utilize the relationship that
we have already derived in the last class,
that is u inner product between u and L v
must be equal to inner product of L star u
comma v plus J u v. So, we write that here,
so what you get is that get inner product
of L star y m and v it was basically y m,
so L star y n, y m plus J inner product of
y m and y n minus inner product of y m, L
y n is equal to, we take integral r y m y
n d x common and this becomes lambda n minus
lambda m.
So, we have already proved earlier that for
Sturm-Liouville problem L is equal to L star
and J u v is equal to 0, so, therefore J y
m y n will be equal to 0, so this will be
equal to 0 in the case of Sturm-Liouville
problem. And what we have now is that inner
product of L star y n and y m minus inner
product of y m comma L y n is equal to integral
r y m y n d x multiplied by lambda n minus
lambda m.
Now, since, L is equal to L star, we can write
as L y inner product of L y n and y m minus
inner product of y m L y n is equal to lambda
n minus lambda m inner product of y m and
y n with respect to weight function r. Therefore,
we have already proved the relationship of
inner product, that is inner product of a
and b should be is equal to inner product
of b. And therefore inner product of L y n
and y m should be is equal to inner product
of y m and L y n.
So, this two will be equal and identical,
they will be cancelling out, so what will
be getting is lambda n minus lambda m inner
product of y m and y n should be is equal
to 0. Now, lambda n and lambda m are two distinct
eigenvalues, therefore lambda n is not equal
to lambda m, therefore to satisfy these equation
only option that is left is inner product
of y m and y n should be is equal to 0, that
means integral a to b y m into y n r d x should
be is equal to 0.
So, therefore, this proves that the eigenfunctions
y n and y m are orthogonal to each other with
respect to the weight function r, so this
proves that eigenfunctions y m and y n are
orthogonal with respect to weight function
r x, this is known as the weight function.
So, this completes the proof that for the
Sturm-Liouville problem, the eigenfunctions
are orthogonal functions with respect to the
weight function r.
Next, we go to theorem number 3. If p of x,
q of x, r of x are real valued functions 
and alpha 1, alpha 2, beta 1, beta 2 are real
constants, these are the coefficients in boundary
conditions, these are real and these coefficients
are 0, then for self-adjoint S L system, the
eigenvalues are real. If the functions the
coefficients functions are real, if the coefficients
in the boundary conditions are real, then
there is no reason that eigenvalues becomes
unreal or imaginary, they will be also real.
So, let us proof this and proof goes like
this, let us assume that eigenvalues are complex,
let eigenvalues are complex, therefore lambda
is equal to C plus i d. So, it has a real
part and it has a complex part, so lambda
is equal to C plus i d, so we have to we write
the eigenvalue problem L y is minus lambda
r y. So, we take so this is equation number
1, so that is the eigenvalue, so that is a
standard Sturm-Liouville problem, we take
the complex conjugate of this equation.
We have already stated that r x is real, so
L y bar, let us say bar is the complex conjugate
of y, so complex conjugate of y is replaced
by the, is denoted by the bar on the top of
it. So, L y bar is equal to minus lambda bar
r, r remains r, because it is a real, that
is our assumption times y bar, so this is
equation number 2, so this is the complex
conjugate.
Now, what we will do, we next we take the
inner product 
of equation 1 with respect to y bar and we
take inner product of equation 2 with respect
to y and subtract one from another. Let us
see what we get, if we really do the subtraction,
you will be getting inner product of y L y
bar y bar L y minus y L y bar may be d x here,
is equal to lambda bar minus lambda integral
r y y bar d x.
Now, we write this equation, we substitute
is as by y L star y bar d x plus bilinear
concomitant between y and y bar minus y integral
y L y dash y bar d x is equal to lambda bar
minus lambda integral r y y bar d x. Now,
since it is a Sturm-Liouville problem, we
have L is equal to L star and J between y
and y bar should be is equal to 0, bilinear
concomitant vanish, as well as the operator
is self-adjoint operator.
So, once we know this facts, then we can simplify
the equation as y L y bar d x plus J will
be equal to 0 minus y L y bar d x is equal
to lambda bar minus lambda integral r y y
bar d x. So, these two quantities on the left
hand side, they are identical to each other
and opposite in sign, so they will subtract,
so lambda bar minus lambda becomes in lambda
bar minus lambda and will be will be taken
out and r y and y bar d x is equal to 0.
So, we have already seen that r is a real
quantity, real function and it is a non-zero
function, so r cannot be equal to 0. So, what
is the product of y and y bar? Product of
y and y bar, this is a complex number you
have said and this is a complex conjugate
of that, if you just do the product, if we
just product two quantities, which is complex
and its conjugate, so this becomes a square
minus b square, so this become a square, i
square is minus 1 plus b square,
So, multiplication of a complex and its conjugate
will be always giving raise to a real part.
Therefore, y multiplied by y bar is nothing
but mode of y square of that, so this is a
real part. What I mean is that the part in
the integral r y y bar d x will be always
real and positive, so this is a real and positive.
So, therefore, in order to satisfy this equation,
only option is left is lambda bar is equal
to lambda. That means complex conjugate equal
to real part, that means the complex part
does not exists, this simply means C plus
i d is equal to C minus i d, so this simply
means that d is equal to 0 that means lambda
is always real.
So, if you have a function p q r, which are
all real functions and coefficients in the
eigen on the boundary conditions beta 1, beta
2, alpha 1, alpha 2 all real, then eigenvalues
of the system or the equation will be always
real, you would not be having a complex eigenvalue.
That means for real system, the eigenvalues
are real and it will be a self-adjoint system.
Now, let us take a stock of whatever we have
done, the summarize. First is, we define the
various classifications of differential equations
of PDEs. Their linearity homogeneity 
And we have already we checked and defined.
Then, we define the principle of superposition
for linear operator, we define the adjoint
operator, we define the characteristics 
of self-adjoint problem, adjoint operator
and then we looked into the properties of
self-adjoint operator.
So, the important properties are number one,
the eigenvalues are real and eigenfunctions
are orthogonal to each other, eigenfunctions
are orthogonal. So, with this background,
we will be in a position to solve the partial
differential equation, linear partial differential
equation by using separation of variable method.
So, in the next class onwards will be taking
up the solutions of partial differential equation.
And up to this class, we are equip with all
the weapons in order to attack, to solve the
partial differential equations by using separation
of variable.
Thank you very much.
