Hello friends welcome to the lecture of integral
equation fredholm integral equation with the
symmetric kernel So if you look at if you
remember we have discussed certain theorem
corresponding to fredholm integral equation
which symmetric kernel So 1st theorem we have
discussed is that if a kernel is symmetric
then all its iterated kernels are also symmetric
And 2nd result which is a kind of hunting
license to start with that is every symmetric
kernel with a nonzero norm has at least one
eigenvalue So this is the beginning point
by which we want to start our theory
So 3rd is that if we have more than one Uhh
more than one Eigen functions corresponding
to Uhh distinct eigenvalues then they are
orthogonal to each other And theorem 4 says
that the eigenvalues of a fredholm integral
equation with the real symmetric kernel are
real okay And next result is which we have
discussed states that the multiplicity of
any nonzero eigenvalue is finite So multiplicity
is the number of linearly independent Eigen
functions corresponding to a given eigenvalue
is always finite provided we have a symmetric
kernel and this quantity is finite
So this we have discussed in previous lecture
now let us utilise the proof of this theorem
5 to prove one more result which says that
the eigenvalues of a symmetric L2 kernel from
a finite or infinite sequence with no finite
limit point It means that when we have a sequence
of Eigen functions if it is finite then Uhh
no problem but if it is infinite then we do
not have finite limit point so they will converge
to infinity So for that we just look at that
suppose we have a sequence say lambda I so
we have a sequence say lambda I and the corresponding
sequence Eigen functions we are denoting as
Phi of x
So without loss of generality we can assume
that all the Phi I x are all orthonormalized
by a gram Schmidt process Now if you remember
in previous proof we have simply approximated
your k of xt as a summation your ai Phi I
x here So I is equal to whatever Ix we have
So here we are taking the summation of over
ai So this is the beginning point here we
are assuming So here I am assuming that U
I bar lambda Uhh I lambda t is your Uhh Eigen
functions corresponding to lambda So if you
drop this notation lambda then you can say
that it is nothing but this
So here we are assuming that k xt is approximated
by ai U I bar t So here U I bar t is the corresponding
so Eigen functions corresponding to your eigenvalues
lambda I is that okay So in the same way you
can define your ai ai is basically a to b
k xt U I t dt So I am dropping this lambda
because we are considering all the Eigen eigenvalues
So if you remember that U I is the Eigen functions
corresponding to say lambda I eigenvalues
So here I am assuming that if we have repeated
say eigenvalues we count them as lambda 1
lambda 2 and so on
So it may happen that lambda 1 lambda 2 may
be equal and same as lambda but we are counting
all the eigenvalues okay So we are saying
that this UI is the Eigen functions corresponding
to lambda I So here ai is nothing but lambda
I inverse U I x So here I am writing here
let me use this notation since I am using
this let us assume this this Phi I is replaced
by your UI So this is U I x is the Eigen functions
corresponding to this lambda I so I can write
it like this So here we have say U I bar because
if this is Eigen functions then these are
also Eigen functions okay
So here I can approximate k xt by this So
here ai is basically what ai is a to b your
k of xt and U I t dt Now this is since by
the property of Eigen functions this is 1
upon lambda I your U I x is that okay So here
ai you can get it like this 1 upon lambda
I UI x Then again we can use Bessel’s inequality
and you can have this property this equation
11 So along equation number 11 you have this
a to b k of xt square dt is greater than or
equal to summation 1 upon lambda I square
and it is what modulus of U I t square this
is X I think so this is x here
Okay so now again now let us again integrate
with respect to x So a to b dx here then here
we have a to b t of x right And we simply
assuming okay This I can consider since this
U I x and all normalised then I can write
this as this is greater than equal to summation
1 upon lambda I square Is that okay So let
me use the okay So here we have this thing
Is that okay Now if this quantity is finite
okay so here is this quantity is finite then
we can say that okay let me write it here
Then this series summation 1 upon lambda I
square is sum is going to be finite
Now sum is going to be finite means you are
infinite series of 1 by lambda I square is
convergent series And if you remember there
is a small result that if a series converges
then its nth term is tending to 0 as n tending
to 0 So it means that this implies that limit
n tending to infinity your 1 upon lambda n
square is basically tending to 0 Or equivalently
we can say that this implies that limit n
tending to infinity your lambda n is going
to be infinity So there are only 2 possibilities
that this sum is finite if this sum is finite
no problem and if this sum is not finite then
we can use this property of convergent series
which says that your nth term is tending to
0 it means that your lambda n is tending to
infinity
So it means that either your so this proves
this thing that the eigenvalues of a symmetric
L2 kernel form a finite if it is finite set
fine or if it is an infinite sequence with
no finite limit point Now this is a rough
sketch of the proof the exact proof proof
of this book is this proof is given in the
book by RP kawal linear integral equation
So now let us move little bit further which
says that theorem 7 it says what let the sequence
Phi nx be all the Eigen functions of the symmetric
L2 kernel k xt with lambda n as the corresponding
Eigen values
So here we have we are able to calculate all
the Eigen values and Eigen functions Then
with the help of this we try to define new
quantity which is known as truncated kernel
Which is what if it is 1st kernel so it is
k1 xt is your k xt and your k2 xt is basically
kxt minus this quantity Phi 1x phi m bar t
lambda m So it is kind of we are approximating
your kernel k xt by the this thing
So here basically we are doing like this Is
that okay So here we try to find out say as
accurate as possible So here we are truncating
I equal to1 1st approximate is 2 3 and so
on So here we are defining that nth plus n
Plus1th place it is given by this So k xt
minus m equal to 1 to n phi mx into some constant
that constant I am writing as phi m bar t
upon lambda m So it is kind of approximation
of k xt with the help of m 1st m Eigen functions
So we call it n Plus1 th truncation of this
k xt So this is known as truncated kernel
Now we can prove that this truncated kernel
has the eigenvalues lambda n Plus1 lambda
n plus2 and so on which is corresponding to
Phi n Plus1 x Phi n plus 2x Eigen functions
So and this kernel will not have any other
eigenvalues or any other Eigen functions So
this is corresponding so it means that if
we already know that eigenvalue is an Eigen
function then we can always find out eigenvalues
an Eigen functions corresponding to this truncated
kernel Now what is the use of this we are
trying to with the help of this we are trying
to approximate your symmetric kernel with
the help of this kind of separable kernel
kind of thing
So here if you remember there is a result
in separable kernel that if we have a separable
kernel then eigenvalues are finite eigenvalues
Now here if I look at that theorem 8 which
says that a necessary and sufficient condition
for asymmetric L2 kernel to be separable is
that it have a finite number of eigenvalues
So it means that this is an if and only if
result that if we have finite number of eigenvalues
then it means that at some point this process
will stop So it means that suppose we have
say we have only n eigenvalues then we can
have say only m Eigen functions
So it means that at kn Plus1th stage this
is nothing but we we we cannot get kn Plus1
xt so it is simply 0 So in that case your
k xt is written as m equal to 1 to n phi mx
phi m bar t lambda m Or we can say that this
is nothing but given in terms of separable
form So if we have only finitely many Eigen
values we can say that in that case your k
xt is given by this separable form Or if I
say that we do not have if it is separable
then we already know that we have only finite
number of eigenvalues So if it is finite number
of eigenvalues then it is separable if it
is separable then we have only finitely many
eigenvalues that we have already done
So here with the help of theorem 7 we can
say that if we have finite number of eigenvalues
then k xt can be written as this kind of form
which is nothing but a separable form for
this k xt So that it means that a symmetric
L2 kernel is separable if and only if we have
finite number of eigenvalues okay So now let
us move to the next result which is a very
very important result of Hilbert Schmidt known
as Hilbert Schmidt theorem It says that again
we are taking this without proof but statement
is very very important
Let us say that that if we have a function
f of x which is given by k xt gt dt we can
say it is generated by a kernel k and a function
g Here we are assuming that this k kernel
k is symmetric L2 kernel and g also we are
assuming that it is L2 function Then Hilbert
Schmidt theorem says that then this function
f of x which is generated by this k and g
can be expanded in an absolutely and uniformly
convergent fourier series with respect to
orthonormal system of Eigen functions psi
1 psi of the kernel k That means that what
it says let me explain in a little bit detailed
manner
It means that if we have f of x is equal to
your k of xt gt dt so here we are assuming
that k xt belongs to L2 and similarly your
gt k and g both belongs to L2 Then you are
Hilbert Schmidt theorem says that fx can be
written as summation of some am phi mx and
summation over this m m is say 1 to infinity
Now what is this Phi m so here phi m is the
Eigen function corresponding to this kernel
k xt Means what that it satisfies this property
that that phi mx is equal to lambda a to b
and k xt k xt phi mt dt
So it means that that this phi mx is basically
Eigen function corresponding to some eigenvalues
for let us say that lambda n and it satisfies
this property And in addition we are just
assuming that this can be orthonormalized
to say psi 1 2 and so on psi 1 psi 2 and so
on So here we can write this as f of x equal
to summation m equal to1 to infinity am psi
mx Now what is the difference between this
and this difference between this and this
is that here it is Eigen functions now by
Gram Schmidt orthonormalization we can convert
into new systems having equal number of elements
here but now with the property that they are
orthonormalized is that okay
So it means that fx can be written as am psi
mx and here am you can write it as fx a to
b and psi mx d of x And just for simplicity
we are writing this as f of we are denoting
this as f of x So if you look at look at equation
number 13 it says that if we have this function
which is given by this then this fx can be
represented as this infinite series which
is absolutely and uniformly convergent So
here f of m is given by this f psi m which
is denoted by this a to b f of x psi mx d
of x So this is the inner product we are defining
as a to b f of x psi m x d of x is that okay
So these coefficients are known as fourier
coefficients here and then this fourier coefficient
fm is related to fourier coefficient corresponding
to g So here I can say that this fm is given
by g of m divided by lambda m where gm is
the fourier coefficient of g and lambda m
are the eigenvalues corresponding to say psi
m and it is eigenvalues of the kernel k This
is not very difficult here so what we can
do here is Uhh to find out this relation 14
fm equal to gm divided by lambda what we can
do here we have this Then we simply multiply
by say phi mx and then use a property of orthogonality
let me write it here we have this okay
So here let us say we have this we have say
f of x phi mx here so here let me write it
f phi m right So this I can write it as f
Phi f is what let me denote this as kg so
this is denoted as defined as k operating
on g so this is kind of an operator on this
So k on g kg is defined as this okay limit
is say a to b so kg Phi of m So this you can
prove that this is same as g k star and phi
m Now since k star is same as k because we
are assuming that it is a symmetric kernel
then it is nothing but gk of phi m
Now we already know that k phi m means this
k xt phi mt dt So this is going to be phi
mx divided by lambda So this is what this
is nothing but g your phi m divided by lambda
is that okay So this is you can write it you
can take out 1 upon lambda m out this is g
phi m and this is nothing but gm we are defining
it like gm gm by lambda m So here is fm this
is nothing but your f of m So fm which is
a fourier coefficient of S with respect to
phi m so fm is given by gm divided by lambda
m
So here you can look at that equation number
14 is valid in this way that fm is given by
gm divided by lambda m is that okay So using
this now let us proceed to solve your fredholm
integral equation of 1st we will try to solve
for 2nd type and then we if we have we will
discuss for 1st integral also So let us try
to solve solution of a symmetric integral
equation of non-homogeneous fredholm integral
equation So please remember if it is homogeneous
then we already have solved that is nothing
but your eigenvalue Eigen function problem
So now let us proceed for solving the non-homogeneous
problem that is y of x equal to f of x plus
lambda a to b k xt yt dt Here I am assuming
that k is L2 kernel so that we can utilise
the theory which we have discussed earlier
Okay Now we assume that this lambda is not
an eigenvalue we will consider the case when
lambda is an eigenvalue but for the starting
point let us assume that lambda is not an
eigenvalue And we are able to solve the homogeneous
problem means we are able to find out all
the eigenvalues corresponding to this symmetric
kernel k
So here we are assuming that lambda 1 lambda
2 all the eigenvalues of the kernel k xt and
this psi 1 psi 2 are orthonormal system of
Eigen function of the kernel k xt So that
we already have enquired So theory says that
you can always do it okay So now using the
Hilbert Schmidt theorem this by x minus fx
is written as lambda k xt yt dt So now you
can use Hilbert Schmidt theorem and say that
y x minus fx can be expressed as this infinite
series in terms of Eigen functions corresponding
to this k xt which is uniformly uniformly
and absolutely convergent here
So this is by your Hilbert Schmidt result
And here we try to find out now this fourier
coefficient that is am So that we know that
am is basically what am is nothing but y x
minus fx psi star mx dx
So here your inner product is defined like
this let me use we already know this that
ym yx yx is equal to f of x plus lambda a
to b k xt yt dt now you can take this out
so y x minus f of x this side and it is lambda
a to b k of xt yt dt And then using Hilbert
Schmidt theorem you can always write it like
this as am and here we are assuming that psi
m psi mx dx and how to find out this am am
is nothing but fourier coefficient corresponding
to this yx minus f of x So that is yx minus
f of x and then U psi m star x dx a to b
So if you calculate this this is what this
you can write it you just take it you separate
these 2 integral So yx psi star mx dx minus
fx psi star mx dx so where psi m star is complex
conjugate of psi mx But if you look at your
previous thing that your fourier coefficient
corresponding to your function f of x can
be written as fourier coefficient of the unknown
function given function gx as this fm equal
to gm by lambda m So here in analogous manner
you can say that am which is a fourier coefficient
of y x minus fx can be written as sorry can
be written as fourier corresponding fourier
coefficient corresponding to this yt
So I can write it here am as lambda ym divided
by lambda so you can equate these 2 things
So when you equate these 2 things you can
get your am and y m So am is the fourier coefficients
here and it is written as lambda fm divided
by lambda m minus lambda this is very easy
I can say that here we have am as ym minus
fm and this is coming out to be lambda y m
upon lambda m So if you will compare you will
get y as 1 minus lambda upon lambda m is equal
to f of m So you can get once your ym is calculated
then you can calculate your am also
So here you can get am as lambda fm upon lambda
m minus lambda y equal to this Okay fine so
this is something we want to find out ym okay
So now I can write it yx as what so y since
look at here equation number 16 so yx minus
fx is equal to n equal to this summation m
equal to 1 to infinity am Am you have already
obtained and psi m is already known to you
so you can get yx in terms of fx plus this
infinite series So you can write it here yx
equal to fx plus lambda I am writing the value
of am So value of am is lambda fm upon lambda
m minus lambda
So I can write it here lambda fm lambda fm
upon lambda m minus lambda psi mx dx Now here
you can utilise the value of fm fm is what
fm is the fourier coefficient of f which is
nothing but this a to b I am not writing the
limit because your interval maybe anything
So here fm is basically what ft psi star mt
dt So using the expression for f of m and
I am using t as the integrable variable because
we are already having x So here fm t I am
writing as a to b psi m star t ft dt
So when you write it here and we already know
that this series is absolutely and uniformly
convergent so we can always interchange the
integral sign and summation sign So we can
write it like this f of x plus lambda n equal
to1 to infinity this thing Now if we denote
this m equal to1 to infinity fm upon lambda
m minus lambda or you can say that if we denote
this gamma xt lambda as this m equal to1 to
n psi m is x psi m star t lambda m minus lambda
and say that it is resolvent kernel then your
solution it can be written as this
So here your solution is given as y of x is
equal to f of x plus your lambda times a to
b gamma xt lambda Uhh your f of t dt where
gamma xt lambda is given as this infinite
series m equal to1 to infinity psi sorry psi
mx psi m star t divided by lambda m minus
lambda okay And this series is absolutely
and uniformly convergent Okay So here solution
is given by this Now if you look at here your
lambda choice of lambda is very very important
because if lambda is one of the Eigen value
then there is gamma xt lambda will not exist
So here I am assuming that the singular point
of this resolvent kernel is the values of
lambda which is equal to lambda m or you can
say that the singular point of the resolvent
kernel gamma corresponding to a symmetric
L2 kernel are simple poles because at the
simple pole every pole is an Eigen values
of the kernel Or you can say that for lambda
equal to eigenvalues is your the singular
point of this kernel gamma resolvent kernel
gamma xt lambda So using this now let us try
to apply the result for actually solving the
Fredholm integral equation of the 2nd kind
okay So that we are going to do it in the
next lecture thank you very much
