We were looking at an example of periodic
Sturm Louisville system what we have seen
is we we have seen that Uhh operator we just
we were trying to find the eigenvalues we
found the eigenvalues and eigen functions
corresponding to lambda is positive So lambda
equal to 0 and lambda as negative that is
lambda equal to minus mu square we have to
check whether they are they there may be any
eigenvalues in them Okay corresponding eigen
functions if there is any eigenvalue okay
So we will try to see those 2 other cases
So we start with lambda equal to 0
If lambda equal to 0 if you do what is equation
said the equation is y double dash plus lambda
y equal to 0 so basically equation becomes
lambda is 0 that is why this is 0 So the general
solution is y of x equal to C1 x plus C2 x
is between a to b okay now you apply the boundary
condition bound that is y at a is equal to
y at b This gives me C1 a plus C2 equal to
C1 b plus C2 C2 C2 cancel so this will give
me C1 has to be 0 because a and b are different
from So C1 has to be 0 if you want this to
be same Okay So what is your general solution
then the general solution becomes C2 just
a constant Okay and now clearly y dash of
a equal to y dash of b here if you apply for
this general solution this will give me 0
equal to 0 satisfied okay
So the eigenvalue 0 is you get what is a solution
so y x equal to constant that is a nonzero
solution If I choose my C2 is nonzero nonzero
that is a solution that is actually satisfying
the equation when lambda equal to 0 and the
boundary conditions okay for them so we choose
C2 equal to1 so is an eigen function because
they have a nonzero solution eigen function
corresponding to corresponding to an eigenvalue
lambda equal to 0 okay So this is simply constant
so which you can see from the earlier case
when Uhh lambda n is equal to 4n square by
pie square by v minus a the whole square and
when I put n equal to 0 if I include 0 here
lambda is 0 that is the 2nd case
What happens to the eigen functions When I
put n equal to 0 this becomes 0 this is 0
and what about you cos 2n pie by Vminus a
into x when n equal to 0 this is simply one
So I already have eigen functions So I can
include this lambda equal to 0 case into the
earlier case by including n equal to 0 So
I can make now n is from 0 1 2 3 onwards I
have these are eigen functions corresponding
eigen functions eigenvalues These are eigenvalues
and corresponding eigen functions are sin
and cosine but the only thing is n is equal
to 0 lambda is 0 lambda 0 is 0 but eigen functions
are not 2 but they are only 1 that is because
sin of 0 sin n equal to 0 it becomes sin 0
is 0 So this function is 0 function
We are looking for only nonzero solution so
eigen function should be nonzero so that is
only that is when we put n equal to 0 into
the cosine function this becomes one so I
already have here Okay I can include this
case into that now we have to see what happens
to this lambda negative okay that is minus
mu square which is the negatives okay So this
is the case if you see the general equation
is given differential equation is lambda square
minus mu square y equal to 0 So its general
solution again if you look for general solution
K square minus mu square equal to 0 K equal
to plus minus mu so you have C1 e power mu
x plus C2 e power minus mu x
So this is negative mu is always positive
okay So that is how it is So you apply the
boundary condition y at a is equal to y at
b if you apply so you get C1 e power mu a
plus C2 e power minus mu a equal to C1 e power
mu b plus C2 e power minus mu b So what you
get the equation is C1 this minus this so
you get e power mu a minus e power mu b plus
C2 Times a power minus mu a minus e power
minus mu be equal to 0 Okay So this equation
number-one if you apply other boundary condition
other periodic boundary condition what you
get is mu into C1 e power mu a minus C2 into
e power minus mu a that is what if you differentiate
and put x equal to a which is same as mu times
C1 e power mu b minus C2 e power minus mu
b
So mu mu you can cancel because mu is positive
so nonzero and you get C1 e power mu a minus
e power mu b and you have minus C2 e power
minus mu a and this when you bring it to this
side it becomes plus and you have finally
minus so minus minus plus minus mu b equal
to 0 so this is equation number 2 If you actually
see this if you substitute so if you want
to have a nonzero solution okay this 1 and
2 you rewrite 1 and 2 actually gives me a
system actually if you are putting as a system
we have a matrix e power mu a minus e power
mu b e power minus mu a minus e power minus
mu b similarly here e power mu a minus e power
mu b minus e power minus mu a minus e power
minus mu b C1 C2 equal to 0 0
So you want to get a nonzero solution here
To get C1 C2 nonzero solution okay this determinant
has to be 0 that is the determinant of a power
mu a minus mu b e power mu a minus e power
mu b e power minus mu way minus e power minus
mu b this determinant minus e power minus
mu a this has to be 0 Okay because it is like
a x equal to 0 to get a nonzero solution if
you are looking for the determinant of a has
to be 0 So this is if you the determinant
means e power mu a minus e power mu b take
it out similarly e power minus mu a minus
e power minus mu b you can take it out from
the 2nd column so what you get is 1 1 1 minus1
which is nothing but this is simply this is
nonzero and this is nonzero then what you
have the determinant is simply the power mu
a minus e power minus mu b e power minus mu
a minus e power minus mu b
This is simply minus1 minus1 that is going
to be minus2 so modulus is minus2 so minus2
This is never be 0 for every mu positive you
can easily see for every mu positive and a
is not equal to b this quantity is nonzero
and this quantity is nonzero so 2 is nonzero
so it is never be 0 for any mu value So you
do not have this determinant is never be 0
okay This determinant cannot be 0 this determinant
is actually nonzero since okay since since
this is actually the determinant is this nonzero
So we see that that means you do not have
mu Uhh if there is no nonzero solution that
means C1 C2 has to be 0
Is C1 C2 both are 0 that means y of x is the
0 solution So completely 0 solution you will
get in this case that implies this mu equal
to this lambda equal to minus mu square for
any mu positive is not an eigenvalue Okay
Implies no eigen function So what you have
finally all the 3 cases lambda positive lambda
equal to 0 lambda equal to minus mu square
you have eigenvalues are eigen functions okay
So you can now put it together all the eigen
functions were eigenvalues eigenvalues 1st
of all I have only lambda n which is for square
n square pie square by b minus a whole square
now for n is equal to 0 1 2 3 onwards okay
eigen functions of you can call them bn of
x and which are cos 2 n pie by b minus a into
x and U n of x which are sin 2n pie by b minus
a into x again n is running from 0 1 2 3 onwards
Okay
So these are your eigenvalues and eigen functions
they are orthogonal okay and they are actually
form complete orthogonal set complete means
I can write any square integrable function
or any piecewise continuous function okay
any piecewise continuous function a piecewise
continuous function this is actually a theorem
so which you will know any piecewise continuous
function fx we will take is fx x is between
a to b okay any piecewise continuous function
so you can have from a to b you can have some
pieces this kind of function okay So any finite
domain that means outside if you want to see
it is like a repeated everywhere like this
So piecewise continuous function that means
you should have a jump discontinuity it is
not continuous but it should not be it should
not have it is ready to not go to infinity
okay at these values it should be jump discontinuity
this limited exist it should be finite okay
as you when you go from this side to this
side the value of the function should be finite
Okay Similarly here so you should have here
and here that means at this point if you take
the limits from this side to the site the
value should be finite Such a piecewise continuous
function I can write as in terms of as a linear
combination of this
I have now running from 0 to infinity what
I have what are the eigen functions I have
an arbitrary constants with these eigen functions
cos 2n pie by b minus a into x plus bn another
set of arbitrary constants corresponding to
with this for this sin 2n pie by b minus a
into x So this I can write fx in terms of
this eigen function and this eigen functions
coefficients are simply arbitrary constants
okay where ans bns are arbitrary constants
okay there are constants How to find these
constants These constants you know you have
not product you can make the dot product fx
with cosine with the bns okay 2n pie by b
minus a into x this you integrate take the
dot product left-hand side that will give
me here a to b an okay cos square 2n pie so
2n pie by b minus a into x
you actually multiply with 2n so let us use
2n okay So that you have what you have is
you multiply cos 2n pie 2n pie cos 2n pie
cos square So corresponding to n equal to
m that will be together cos square 2n pie
by ba b minus a Other things will be cos 2n
pie by b minus a into x into cos 2n pie by
b minus a into x That integration the dot
product is 0 because n is not equal to m only
n equal to m that is cos square 2n pie by
b minus a into x into an this is what you
get So and if you do for sin you get the same
thing So you do cosine or sin you get the
similar thing
So you get sin here instead of cos you get
here bn instead of an Okay So this is running
from including m is m is running from 0 1
2 3 onwards okay Of course b0 is 0 a 0 only
will contribute a0 will be nonzero okay What
is the reason because m equal to 0 sin square
that is 0 0 into that is 0 right 0 divided
by here also seen 0 0 by 0 that is actually
what you have is bn is anyways 0 bn into 0
whatever bn maybe arbitrary arbitrary constant
So we can always write I am simply writing
corresponding to n equal to 0 to infinity
for cosines sines from 1 to infinity and that
0 part you can write like the 0 into 0 so
it does not matter so b0 is arbitrary it is
anyways 0
So you can have this kind of expression okay
So to write it separately so a to b fx sin
2n pie by b minus a into x equal to bn a2
b so you have now sin square sin square 2n
pie by b minus a into x dx So you should not
forget this dx So this is this is what you
get for n is running from So this will give
me those fourier coefficients an equal to
you can now calculate this integral these
integrals you can calculate that is one plus
cos 4n pie by b minus a into x okay by 2 right
This sin once you put sin that will become
0 again like earlier we have seen cosine at
a minus cosine at b that is 0 Okay
Cosine at b 4n pie by b minus a into b minus
cos for n pie by b minus a that is 0 So this
will not contribute finally what you get is
an is 1 divided by what you get is here so
this integral value is 1 by 2 integral a to
b so that is b minus a by2 So what you get
is if you bring it to the other side an will
be 2 divided by b minus a this integral a
to b fx cosine 2n pie by b minus a into x
dx this is known okay and similarly you get
bn as again you get the same thing instead
of plus you have a minus that is 2 sin square
right So this is again so this will not contribute
when you evaluate and so what you get is the
same so you get 2 divided by b minus a integral
a to b fx sin sin 2n pie by b minus a into
x dx So these are your fourier coefficients
Given a signal time signal fx you can have
these frequencies discrete frequencies okay
with these amplitudes and you and you can
combine it So these are actually your fourier
transforms fourier transforms Given a time
signal fx you can have these fourier transform
coefficients and then to get back your signal
using these frequencies using the discrete
frequencies you combine is a fourier series
this is what is the fourier series from the
fourier series you can get back your signal
based on discrete frequencies Combine all
the discrete frequencies you can get back
your fourier series Okay
So this is why this is what you have seen
So now you can use anything so what you might
see is a is 0 b equal to 2 pie is a regular
fourier series what you see in the text books
okay 0 sometimes some others may write minus
pie b equal to pie these are this okay but
taking this what you get is regular fourier
series fourier series of time period 2 pie
Otherwise general fourier series between with
the period b minus a you can have this fourier
series okay general fourier series So this
is how regular periodic Sturm Louisville system
will give you your fourier series regular
fourier series which you study in the engineering
So what we have seen is a regular Sturm Louisville
system also gives you some kind of fourier
series and this periodic Sturm Louisville
system is actually giving the series fourier
series which you studied in the in your engineering
Okay So what you actually study is this periodic
system periodic system Sturm Louisville system
that gives the fourier series now let us see
the 3rd case singular Sturm Louisville system
okay I will give you an example of singular
Sturm Louisville system that from which for
which if you find eigenvalues and again functions
you can get a fourier series there as well
So there I will try to see the example in
the 3rd type singular Sturm Louisville system
example of singular Sturm Louisville system
I do not really do something Uhh new here
so which already know I will just give you
an example of 2 examples of 2 differential
equations examples of 2 differential equations
which you have already studied One is Legendre
equation other one is Bessels equation that
is what we study here in the we will give
an example here for the singular Sturm Louisville
system So let us take this Legendre equation
Legendre equation what is the Legendre equation
If you remember 1 minus x square y double
dash minus2 xy dash plus alpha into alpha
Plus1 into y equal to 0 So this is what is
our Legendre equation right So this is equation
it is defined between minus1 to1 and you see
you can rewrite this put it like Ly equal
to lambda y like self adjoint form or skew
symmetric form where L is skew symmetric if
you put it in this form what is my L L equal
to is actually Uhh 1 1 by minus 1 is 1 by
W is 1 so what you have is P is 1 minus x
square d dx of okay and d dx of this whole
thing So you have ddx of 1 minus x square
into ddx Okay
and Q is 0 Q is 0 so you have there is no
Q here this is your L what is lambda lambda
is simply alpha into alpha Plus1 Okay So 
because what is P that means Px is 1 minus
x square which is actually 0 0 at x equal
to1 or minus1 That means it is a singular
self adjoint equation singular Sturm Louisville
type of equation So boundary conditions should
be what are the boundary conditions to prescribe
to p to p what are the boundary conditions
to prescribe in this Uhh self adjoint singular
Sturm Louisville system When both P of a P
of a is P at 1 and P minus1 both are 0 boundary
conditions should be y of x rather y at a
here y is 1 y at 1 y dash at 1 are bounded
They are finite similarly y at minus1 and
y dash at minus1 are finite So actually minus1
is a smaller one so a is this b is simply
one So these are the finite bounded R finite
So these are the boundary conditions So you
have already studied when do you have your
bounded solutions you have it has a solution
Pn of x these are the bounded solutions bounded
solutions for the Legendre equations all other
solutions are series solutions which are unbounded
plus minus1 So the bounded only bounded solutions
are Pn of x That means lambda is equal to
when lambda is n into m Plus1 okay I have
a corresponding solution Pn of x which is
nonzero Okay
These are the eigenvalues corresponding to
this lambda I have a nonzero solution Pn of
x which is satisfying the boundary conditions
Okay So you call this Vn of x as Pn of x and
this you call as lambda n So these are eigenvalues
and these are corresponding eigen functions
They that into check okay lambda is positive
negative equal to 0 all those things are already
checked verified what are the solutions okay
So your eigenvalues and again functions are
these in this case okay
Immediately implies what is the dot product
so you can see the dot product so there is
no W is 1 so the dot product is you can you
can also write the dot product fg is actually
integral minus1 to1 because determinant is
from minus1 to1 fx gx bar dx Okay So so for
the sake of completeness you can write lambda
n or n into n Plus1 eigenvalues corresponding
eigen functions are Vn of x which are Pn of
x eigen functions corresponding to n is from
0 1 2 3 and so on Okay
but n is from 0 1 2 3 onwards you have this
P0 P1 P2 and so on What happens if I take
Pn equal to minus1 If n equal to minus1 I
have still P minus1 of x this is also polynomial
okay What is this actually this is actually
you have shown that it is actually minus1
power 1 okay minus1 power minus1 into P1 of
x you have seen P n of P minus n of x equal
to minus1 power n into Pn of x Using this
relation which we proved earlier so we can
say that P minus1 and P1 they are actually
linearly dependent So corresponding to this
you have same solution
So if if we think that this is your eigenvalue
okay if n is corresponding to n equal to minus1
so what is that eigenvalue you will lambda
minus1 which is equal to minus1 into 0 So
this is actually lambda 0 which you already
have here okay If you take lambda equal to
minus2 okay lambda equal to minus2 so if you
write lambda equal to minus2 so if you think
that lambda minus2 is also an eigenvalue lambda
minus2 because you have p minus2 okay P minus2
is P2 so which is nonzero nonzero solution
I have eigen functions but I want to see whether
this lambda mod is lambda minus2
Lambda minus2 is minus2 into minus 2 plus1
into 1 minus1 This is nothing but simply 2
right minus2 into minus2 Plus1 so this is
simply minus2 so this is 2 2 is corresponding
to lambda equal to1 so lambda 1 So lambda
1 is already here corresponding to n equal
to1 okay Lambda 1 is also 2 so like this all
lambda minus ns they are already here they
are same same eigenvalue okay So let us not
bother about this lambda is negative negative
natural number If lambda is minus n okay if
we choose n equal to negative values negative
integers you the eigenvalues are already here
you do not have to they are not they are not
different from these eigenvalues that is what
I mean to say okay
So you can say as a note if since lambda ns
lambda minus ns or 0 1 2 3 there actually
00 lambda 1 lambda 1 lambda 2 and so on for
n is from 0 1 2 3 onwards Okay Lambda minus
n for if you choose lambda minus n n is from
1 2 3 onwards okay you can rewrite So lambda
n where n is from 1 2 3 so if you consider
this nothing but lambda 0 lambda 1 lambda
2 and so on these are here 
So you can say note if lambda n is this for
lambda ns these are 
I should properly I should write for n for
n is equal to 1 2 3 onwards lambda minus n
are in one of these eigenvalues
So we need not consider negative discrete
numbers so negative integers So you have these
ones so once you have this if you use the
properties of these eigen functions self adjoint
operator or skew symmetric operator any skew
symmetric any any piecewise continuous function
piecewise continuous function fx I can write
x is between minus1 to1 I can write in terms
of Cn Pn of x n is from 0 to infinity that
is what we have So where Cn is how do I find
my Cn you take the dot product with fx With
P n equal to you get Cn times Pn Pn all other
things will be 0 okay So that means Cn is
you can rewrite so you can write integral
minus1 to1 fx Pn of x these are all real valued
so there is no bother does not matter
So divided by minus1 to1 Pn square of x dx
you know the value of this this is actually
2 divided by 2n Plus1 so you have 2n Plus1
by 2 okay So this value is 2 divided by 2n
Plus1 okay So P0 is 1 P0 is 1 1 is simply
one is 2 right so 2 divided by 2n Plus1 is
the integral of minus1 to1 dx is equal to
2 okay One Plus1 so 2 so 2 divided by this
is actually 2 divided by 2 into 0 Plus1 So
2 so this value is 2 divided by 2n Plus1 So
which we know So this is equal to 2n Plus1
divided by 2 integral minus1 to1 fx into Pn
of x dx This is my Cn
So I have a fourier series bessel this is
called Legendre fourier series and these are
fourier Legendre fourier transform you can
say these are fourier transforms Legendre
fourier transform So given a time signal I
can split I can make it discrete frequencies
with these functions P on and get the discretes
call discrete means 0 1 2 3 onwards okay So
I can have these frequencies and I can combine
with these functions instead of sines and
cosine I can have these Legendre functions
I can combine them discreetly
I can combine is the discrete sum with these
discrete frequencies I can combine them as
a linear combination I can get back my signal
fx this is what is fourier series is all about
So you can have this Legendre fourier series
if you consider this singular Sturm Louisville
system okay So why I choose your piecewise
continuous function So if I do this any fx
I can write in terms of this So that means
this sum is converging to fx you fix any value
x that is that is point wise convergence that
is called quite wise convergence
That means the series you fix x the convergence
is to corresponding fx okay This has a series
of numbers once you fix x the series converges
to Sx at that point x okay Once you fix x
this is point wise convergence but there is
a but in all the 3 cases if f of x is that
means regular Sturm Louisville system or periodic
Sturm Louisville system all these singular
Sturm Louisville systems if fx is a square
integrable function square integrable function
then then this fx still I can write this fx
as sum n is from 0 to infinity So for example
just for the sake of example I can do in this
case Legendre fourier series I can write this
Pn of x where Cns are same here okay
Cns are same but the convergence here is not
point wise not point wise convergence That
means this convergence means I take n is from
1 to n Cn Pn of x okay minus fx okay you director
minus1 to1 okay This square dx equal to 0
when you take this limit limit m goes to this
limit goes to 0 That means as in the square
integrable sense on an average this converges
to fx okay that is this okay So this convergence
is different so if it is not quite wise convergence
but it is a square integrable convergence
it is called square integrable convergence
then this means this is same as this means
not you fix your x and then see that this
number series converges to the particular
value okay
It is actually is this meaning of this one
so this this you need not worry so that is
why you learn only piecewise continuous functions
you can have this fourier series it is point
wise convergence okay but square integrable
square integrable convergence that is this
one this is the meaning of square integrable
convergence This is true even in the regular
Sturm Louisville system or periodic Sturm
Louisville system which I missed to explain
last time okay So if you take square integrable
function fx then still this fourier series
converges to that f but in the square integrable
convergence okay
but if you take piecewise continuous function
fx this convergence this series is converging
to fx point wise okay We have studied bessel
bessel equation okay earlier So we can give
operator as a bessel equation operator and
we can have an operator of a singular Sturm
Louisville system another example that is
bessel equation bessell Sturm bessel Uhh Sturm
Louisville system So the operator L is the
bessel type so that we will see that example
in the so we can give one more one more singular
Sturm Louisville system with certain boundary
conditions okay So we will see that in the
next video thank you very much for watching
this
