Hello friends welcome to the lecture of Integral
Equation and Calculus of Variations with its
applications and examples we were considering
the case of symmetric kernel
So if you remember we have we were doing this
problem y of x equal to f of x plus lambda
a to b K of x t y of t y of t d of t provided
that this K of x t is equal to K bar t of
x So here we are assuming that we have a symmetric
kernel and then we try to solve the fredholm
integral equation of second kind with this
thing So for that we have already discussed
theory corresponding to eigenvalues and eigenfunction
and how to solve this kind of equation let
us apply that theory to solve our problems
So our first problem is this y of x equal
to x plus 1 whole square plus a to b here
a is minus 1 and b is 1 xt plus x square t
square y t dt Now if you look at here your
f of x is your x plus 1 whole square and lambda
is basically 1 here and if you look at K x
t is this so here K x t is same as K t x here
we are considering real so bar is not playing
any role So K x t is equal to K t x for this
particular example
If you remember this is of one more type that
is separable type so you can solve with the
help of separable type also but we are solving
this with the help of theory which we have
developed earlier So here we first we need
to find out say eigenvalues and eigenfunction
then to find out eigenvalues and eigenfunction
let us consider the homogeneous version homogeneous
version means this f of x is not there so
we are considering this homogeneous problem
that is y of x equal to lambda minus 1 to
1 xt plus x square t square y t dt and we
want to find out the values of lambda for
which we have a nontrivial solution and those
that values those values we are saying that
it is eigenvalues and corresponding solution
eigenfunction
So we are solving this with the help of method
which we have developed for separable kernel
so we are writing this as lambda x minus 1
to 1 ty t dt plus lambda x square minus 1
to 1 t square y t dt Now this quantity is
something which is a constant call it C 1
and this quantity minus 1 to 1 t square y
t is another constant which we call as C 2
so y x is written as lambda C 1 x plus lambda
C 2 x square so solution we already know only
thing is we need to find out this C 1 and
C 2 So for that we use the expression for
C 1 and C 2 and use the value of y of x
So here we have C 1 is equal to this and C
2 is equal to this so using y t given by 21
you can get two equation 23 and 24 C 1 1 minus
2 lambda by 3 plus 0 times C 2 equal to 0
and this thing Now if you remember by solving
this is nothing but algebraic equation (D
lambda C 1) this D lambda of C 1 and C 2 equal
to this thing So here your D lambda is the
coefficient matrix of this algebraic equation
So D lambda is coming out to be modulus of
1 minus 2 lambda 3 0 0 1 minus 2 lambda by
5 and if you look at this is a simply a triangular
matrix and here you can get your eigenvalues
very easily and you can say that lambda 1
equal to 3 by 2 and lambda 2 is equal to 5
by 2 is the eigenvalues means for which your
D lambda is having 0 determinant So here eigenvalues
we know then we need to find out the corresponding
normalized orthonormalized eigenfunction
So for that let us start with first that is
lambda equal to lambda 1 that is 3 by 2 and
try to solve this 23 and 24 So when you put
lambda 1 equal to 3 by 2 here in 23 and 24
then you look at that the first equation 23
is simply redundant because this quantity
is 0 so 0 times C 1 plus 0 times C 2 equal
to 0 so this is simply will not give anything
So if you look at here by putting this you
can get that C 2 equal to 0 and C 1 is completely
arbitrary
So when we have C 1 is completely arbitrary
then you can write down our solution y 1 x
here we can put the value of C 2 so C 2 is
0 here so y x is lambda C 1 x So I can say
that lambda is basically what lambda is 3
by 2 here so I can write y 1 x is equal to
3 by 2 C 1 x So corresponding to this eigenvalue
lambda 1 equal to 3 by 2 we have a nontrivial
solution that is 3 by 2 C 1 x
Now with C 1 is some arbitrary constant so
we can always take C 1 in a way such that
this 3 by 2 C 1 is 1
So we can say that your y 1 x equal to x is
the eigenfunction corresponding to lambda
1 equal to 3 by 2 Similarly we can find out
(eigen) eigenfunction corresponding to 5 by
2 But we need normalized eigenfunction so
first let us normalize this y 1 x so normalize
means you just divide by norm of this so what
is the norm here so norm here is minus 1 to
1 y 1 x square d of x
Now if you calculate y 1 x is x here if you
calculate this quantity is coming out to be
this 2 by root 6 so you just calculate and
divide by this so Psi 1 x is given by this
root 6 by 2 x So this is very easy to solve
okay So now so lambda 1 is known to you Psi
1 is known to you now repeat the same procedure
for lambda 2 so lambda 2 is 5 by 2 So first
equation gives you this and second is again
the same way redundant
So here you can solve and you can say that
C 1 is coming out to be 0 and C 2 is arbitrary
Again using the solution this 21 C 1 is simply
0 so y x is nothing but lambda C 2 x square
So here lambda is 5 by 2 so it is y 2 x is
5 by 2 C 2 x square Now again in a same way
C 2 is completely arbitrary so we can always
consider C 2 as 1 and so y 2 x is your x square
is the eigenfunction corresponding to eigenvalue
5 by 2 Now you normalize it so when you normalize
it you divide by norm of this so norm is minus
1 to 1 y 2 square dx so y 2 x is x square
simply put it here and then you can find out
the Psi 2 x under root 10 by 2 x square
So once we have Psi 1 and Psi 2 then you can
use our formula which says that our solution
I am just going back to this and here we say
that the solution is written as y x equal
to f of x plus lambda m equal to 1 to infinity
f m upon lambda m minus lambda Psi m x here
Here m is from 1 to infinity but here in our
particular example we have only two eigenvalue
so this infinite sum is reduced to only finite
sum that is m equal to 1 and 2
So what we want We already have lambda 1 and
lambda 2 and Psi 1 and Psi 2 we have only
thing we have to calculate is f 1 and f 2
So to calculate f 1 and f 2 f 1 and f 2 are
fourier coefficient corresponding to f so
(f of) so f 1 is basically what Minus 1 to
1 f of x Psi 1 x dx so Psi 1 x is already
known to you you put it here and f x is already
known to you so you can get the value of f
1 that is 2 root 6 by 3 this quite simple
calculation you can just plug in and then
you can find out f 1
Similarly in a same way you can find out f
of 2 f of 2 is minus 1 to 1 f of x Psi 2 x
so this is the fourier coefficient corresponding
to this Psi 2 So again you can put it all
values already you know f you know Psi 2 you
can get your f 2 f 2 is coming out to be 8
root 10 by 15
So now you know lambda 1 lambda 2 Psi 1 and
Psi 2 and f 1 and f 2 you just write down
your solution as y x equal to f x plus lambda
m equal to 1 to 2 f m upon lambda m minus
lambda Psi m x so here lambda is 1 we have
taken as 1 so f x is this f 1 is this f 1
Psi 1 x upon lambda 1 minus 1 plus f 2 Psi
2 x upon lambda 2 minus 1 so putting all the
values we have y x is equal to 25 by 9 x square
plus 6 x plus 1
So let us consider one more example to have
a better understanding I hope this is not
very difficult Let us look at this problem
y x equal to 1 plus lambda 0 to pi cos of
x plus t y t dt Now here again you look at
cos of x plus t is a symmetric kernel so it
means that if you interchange it is first
of all real then if you interchange the role
of x and t then cos of x plus t is same as
cos of t plus x So it is case of symmetric
kernel and then we try to find out
So here f of x is simply 1 and since lambda
is not given any constant value so we can
say that lambda is same as lambda So to find
out the solution in terms of resolvent kernel
or using our theory we need to calculate our
eigenvalues and eigenfunction So consider
the homogeneous equation that is you take
simply f x as 0 so consider this y of x equal
to lambda 0 to pi cos of x plus t y t dt now
we want to find out say eigenvalues and eigenvectors
So this is this you can solve using the method
of separable kernel so you can find out eigenvalues
and eigenfunction with the help of method
we have discussed for separable kernel So
here you write down cos of x plus t as cos
of cos x cot t minus sin x sin t y t dt Now
again you simplify then it is 0 to pi cos
t y t dt minus lambda sin x 0 to pi sin t
y t dt Now this quantity is again constant
call it C 1 this quantity is your constant
call it C 2
So it means that your y x is already known
to you so this is lambda cos x C 1 minus lambda
sin x C 2 So only you need to find out this
C 1 and C 2
So so here using this we already know that
C 1 is 0 to pi cos t y t dt and C 2 is 0 to
pi sin t y t dt So C 1 and C 2 and you already
know the solution y like this so from this
you can calculate y t put it back to C 1 and
C 2 you can have these two equation C 1 2
minus lambda pi plus 0 times C 2 equal to
0 and this 32 equation number 32 and this
is again if you look at this is again a triangular
equation to solve and here your eigenvalue
can be find out by say coefficient matrix
determinant of coefficient matrix and determinant
of coefficient matrix is nothing but D lambda
and when you solve this you will get that
we have two eigenvalues that is lambda 1 is
equal to 2 by pi and lambda 2 equal to minus
2 by pi
So eigenvalues is with us now we try to find
out the corresponding eigenfunction So let
us start with lambda equal to lambda 1 that
is 2 by pi so when you put lambda equal to
lambda 1 that is 2 by pi then first equation
is simply redundant and you can solve this
and you can say that C 2 is coming out to
be 0 and C 1 is arbitrary as we have considered
in the previous case
So here we say that C 2 is 0 and C 1 is arbitrary
and which says that look at equation number
29 So y x is already given so using C 2 equal
to 0 and C 1 arbitrary y 1 x is equal to 2
by pi C 1 cos of x Again in a same way which
we have already discussed you can take 2 by
pi C 1 as 1 C 1 is arbitrary
So y 1 x is your cos of x is the corresponding
eigenfunction so just normalize it So how
to normalize it you divide it by norm norm
here is a to b a is what 0 and b is pi so
I think it is already cleared here a is 0
and b is pi so 0 to pi y 1 x square d of x
half so this is a norm of y 1 so divide by
this and this quantity is simply pi by 2 so
this is what 2 by pi root of 2 by pi cos of
x
So this is the eigenvalue normalized eigenfunction
corresponding to lambda equal to pi by 2 Now
let us do the same procedure for lambda 2
equal to minus 2 by pi and you can say that
in a same way you can get C 1 equal to 0 and
C 2 is arbitrary and you can write down the
corresponding nontrivial solution as y 2 x
and it is minus of minus 2 by pi C 2 sin x
again you can take this constant as 1 and
y 2 x is nothing but sin of x is the corresponding
eigenfunction
You can again normalize in a same way so Psi
2 x is y 2 x divided by norm of y 2 x which
is nothing but this quantity So please I hope
we already know norm of y of x is equal to
a to b y of x square d of x 1 by 2 okay by
the way this is this is a dummy variable so
you can use t or anything I am using t because
I am using x here okay So this is norm of
y of x so that is why we are dividing by norm
of this So here we have Psi 2 x is given to
you
Now in a similar way we can find out f 1 and
f 2 which is fourier coefficient of f corresponding
to these eigenfunction Psi 1 and Psi 2 so
f 1 is 0 to pi a to b so a to b means 0 to
pi f x Psi 1 x dx and you can say that it
is coming out to be 0 and f 2 is 0 to pi f
of x Psi 2 x dx which you can calculate it
is coming out to be this quantity 2 upon 2
by pi square root of this 2 into square root
of 2 by pi So f 1 f 2 is known to us
Now you simply put it back to your solution
and now problem is that this lambda is not
given any constant value So your solution
will also depend on the value of lambda because
lambda we can take any constant so right now
we have a choice that lambda is none of the
eigenvalues one of the eigenvalues you can
take this thing
So let us consider the case where lambda is
say taking no values taken as lambda 1 and
lambda 2 So lambda is neither lambda 1 or
lambda 2 So in this case you have y x equal
to f of x plus lambda m equal to 1 to 2 f
m lambda m minus lambda Psi m x So writing
f 1 and f 2 you can get the solution like
this Now if you look at here look at this
2 plus lambda pi so if so in case when lambda
is neither of these eigenvalues you can write
y of x as this
Now in a second case when lambda is equal
to lambda 2 so if you look at lambda equal
to lambda 2 here then this is something which
creates problem Now here lambda is lambda
2 means denominator is 0 and if you look at
numerator lambda is some fixed quantity now
look at this Psi 2 is nonzero so only thing
depend on this f 2 norm Now if f 2 is again
0 if f 2 is coming out to be 0 then it is
kind of 0 by 0 form right
And in this case this can take any value So
right now we say that f 2 is nonzero which
we have calculated by equation number 34 that
f 2 is nonzero So in this case it is (0) nonzero
divided by 0 so this cannot happen So it means
that in this case when lambda is equal to
lambda 2 that is minus 2 by pi we do not have
any solution So it means that no solution
exists if lambda has value minus 2 by pi
Now let us take lambda equal to lambda 1 that
is 2 by pi now in this case f 1 is coming
out to be 0 that is calculated here 33 so
it means that here your f 1 is 0 denominator
is also 0 so it is kind of 0 by 0 form So
this I can take any value A so if this limit
exist we call it any value 0 by 0 is you can
take any value A you can take any value so
here we have infinitely many solution so I
can write it A Psi 1 x plus lambda upon lambda
m minus lambda f m Psi m x So here lambda
m is your 2 okay
So here I can write A is any arbitrary constant
and this lambda m is lambda 2 so you can put
it the value here and you can say that it
is 1 plus C cos x minus 2 by pi sin of x where
C is this and since A is arbitrary C is also
an arbitrary thing So in this particular case
if lambda is one of the eigenvalue and if
the corresponding fourier coefficient is nonzero
then we have no solution as we have discussed
in case 2
But if lambda is one of the eigenvalue and
the corresponding fourier coefficient f 1
say f 5 whatever is 0 so it means that when
lambda takes eigenvalues and the f nonhomogeneous
term is orthogonal to the corresponding eigenfunctions
in that case we have infinitely many solution
and this we can see in a this we have already
discussed in the case when we have separable
kernel So in case of separable kernel this
case we have already discussed for the example
when we have taken kernel K x t as 1 minus
3 x t
So here we are considering the same case that
lambda equal to lambda 1 so lambda is taking
the first eigenvalue and f 1 is 0 means that
f is orthogonal to the first eigenfunction
corresponding to the eigenvalue here So if
it is the case then we have infinitely many
solution and if that lambda is lambda 2 and
f is not orthogonal to the second eigenfunction
then we do not have any solution So we have
these many cases
Now with the help of this now let us consider
this fredholm integral equation of the first
kind Here again we use Hilbert Schmidt theorem
which says that f we have f x equal to K x
t is g t dt then where K and g both are L
2 functions Then f x can be written as your
linear combination of eigenfunctions corresponding
to K x t so that we already so Hilbert Schmidt
theorem we are assuming
So here let me write it here so here we are
using this is the Hilbert Schmidt theorem
that f of x is equal to a to b K of x t g
t dt and if K and g both are L 2 function
then f can be written as summation of ai ai
phi i x and summation over i and where phi
x is normalized eigenfunction corresponding
to this kernel K So it means that we are solving
this y of x is equal to lambda a to b K of
x t and y x so here phi i x is the solution
of this so let me write it here phi x lambda
i okay
And this fourier coefficient here f i is related
to fourier coefficient g i like this so here
you can say that your f k is written as g
k divided by lambda k this we discussed earlier
So here I can write g k as lambda k f k Now
if you look at unknown function we are assuming
in the first kind we are assuming that this
g is the unknown function which we are discussing
So how to construct this unknown function
from the fourier coefficient of this unknown
function g
So here we have to recall the Riesz-Fischer
theorem which says that if we have say this
g k right So if this series is absolutely
and uniformly convergent then we have a function
g ofcourse in L 2 such that this g has this
representation g x is minus the summation
this a k sorry this has this representation
g k and phi k x K is equal to 1 to n is tending
to 0 okay or you can say that that g has this
representation okay
So so you can write that if this infinite
series if this infinite series if it diverges
then we do not have any solution but if it
converges then Riesz-Fischer theorem says
that your solution g x can be written as this
limit of this infinite series okay and this
limit this Riesz-Fischer theorem says so So
it means that everything depend on this infinite
series
So let us take one simple example here so
let us take this example here f of x equal
to 0 to 1 K x t g t dt where K x t is given
by this so here it is the fredholm integral
equation of the first kind Then we need to
find out that this is falling in this category
I need to find out f k and g k all these thing
So eigenvalues we need to find out and eigenfunction
we need to find out
So for that corresponding to this we need
to find out the eigenvalue and eigenproblem
So here for that we need to find out we need
to solve this where K x t is given by this
this thing
So to find out this we can let me write it
here phi of x equal to lambda here it is 0
to 1 K x t and your phi t dt okay So from
this you can easily say that by conversion
of this integral equation into corresponding
boundary value problem we can say that the
corresponding boundary value problem is this
d 2 y by dx square plus lambda y equal to
0 with the boundary condition that y of 0
equal to y of 1 equal to 0 so that is clear
from this equation that y of 0 when you put
it here so y of 0 equal to y 1 equal to 0
So to solve this we have to solve this particular
problem so I think this is quite simple problem
and you can say that in this case your lambda
i are simply what pi square and basically
it is n pi square n pi whole square and the
corresponding eigenfunctions are root 2 sin
pi x root 2 sin 2 pi x and you can say that
root 2 sin n pi x are your eigenfunction So
it means that corresponding to this kernel
look at the eigenvalue problem that is g of
x equal lambda 0 to 1 K x t g t dt
And you look at that this is equal and to
the boundary value problem here so find out
the solution here So whatever be the solutions
here we have this solution here and the values
for which we are getting a nontrivial solution
here the same will be nontrivial solution
here So it means that solving this problem
is equivalent to solving this problem So here
we are finding the lambda for which we have
a nontrivial solution
So lambda is coming out to be n pi whole square
and the corresponding eigenfunctions are this
So now you can calculate your fourier coefficient
f k like this root 2 0 to 1 sin k pi t f t
dt Now once we have f k you can calculate
f k and lambda k square Now this is this quantity
pi to power 4 k equal to 1 to infinity k 4
f k square Now if this series converges then
you can write down your g as this limit so
once this lambda f k this series is convergent
then you can write yours g x as limit of this
series
So it means that we need to find out f k such
that this series is converges one this series
is converges you can find out your solution
g is it okay So we will discuss some more
problem in our exercise so thank you for listening
us we will meet in next lecture thank you
