Hello Friends today in this lecture we’ll
going to discuss Fredholm integral equation
with the separable or degenerate kernels so
this is the topic of the today’s lecture
so just recall some basic things that an equation
of the form alpha x y of x equal to f of x
plus lambda a to b k (x t) y (t) dt
This equation unkown function y x is also
coming in under the sign of integration and
outside this kind of integration is known
as integral equation
And if the limit is constant we say that it
is a Fredholm integral equation so here we
are assuming that this alpha f and k are some
given function and lambda a and b are some
constant value and we call this equation one
as Fredholm integral equation and this y (x)
is unknown function which we are going to
find out and the given function k x t which
depend upon the variable x and t is known
as the kernel of the given integral equation
On the basis of this function alpha of x here
we can categorize this integral equation into
three kind first is when this alpha x is (01:36
directly) equal to zero this equation one
is known as Fredholm integral equation of
the first kind so it means that here this
term is simply vanish and we say that f (x)
plus lambda a to b k (x t) y (t) d t is equal
to zero
We call this as Fredholm integral equation
of first kind and if this alpha x is an ideally
equal to one The constant value one then this
is known as Fredholm integral equation of
the second kind and if alpha is not a constant
value but is a function of x then this integral
equation is known as Fredholm integral equation
of the third kind and generally in practice
we discussed only the first and second kind
generally in a very less occasion we discussed
the Fredholm integral equation of third kind
The reason we can say that if we can take
alpha x is positive Or we can say same sign
without lots of generating we are assuming
that alpha x is positive through the interval
a and b then you can take we can simplify
our integral equation which is convertible
into Fredholm integral equation of second
kind just look at here
What we try to do here we have alpha of x
y (x) is equal to f of x here plus a to b
is limit k of xt and here we have y of t here
So what we do here since alpha x we are assuming
that it is all the time positive for all x
between this interval a to b then we can write
it here as alpha x and y of x here divide
by root of alpha x which we can do f x of
root of alpha x here plus a to b and I can
take inside because this integration is with
respect to t so under root alpha of x here
and here I can divide by alpha of under root
alpha t here and under root alpha t we can
write it here y of t d of t
So if you look at if we assume this as new
variable y of x then it is given as y of x
equal to f of x plus a to b k of x of t divided
by root of alpha x root of alpha t here and
this is y of t d of t so it means that if
we assume that your alpha x is having the
same sign or we can say without loss will
generate that it is positive throughout the
interval then Fredholm integral equation of
third kind maybe converted into Fredholm integral
equation of the second kind
So this is the reason why we discuss Fredholm
integral equation of the second kind in mode
of the discussion so now let us consider right
now a very simple case of kernel kexec we
take kernel kexec in this form that we take
k (x t) here as separable or degenerate kernel
So here we are considering this Fredholm integral
equation of second kind when the kernel is
given in separable or degenerate form so what
is separable and degenerate form
So we can simply say that a kernel k x t is
called separable or degenerate if it can be
expressed as summation of finally many terms
which is written in terms of function of x
into function of t here so here I am assuming
that k x t is given as here a i x is a function
of x only and b i t is a function of t only
and I am writing here k x t as summation of
these and terms I equal to one to n a i x
b i t and here we can assume that all these
a i x are linearly independent
In a same way we can say that all b i t b
one t to b n t are linearly dependent if it
is not then we can use into lesser sum of
the similar kind of event so if we say that
if these are not linearly independent so it
means that one of the it can be written in
terms of other pies so for example if i assume
that a one to a n x are linearly dependent
so it means one of a i a k x can be written
as been a combination of other so we can reduce
this sum to lesser say summation ok
So that’s why without loss of generality
we are assuming that whatever present here
is representing the linearly independent function
so here A one to a n’s are linearly independent
similarly b one two b n’s are linearly independent
so when we have this separable kernel then
how we can solve Fredholm integral equation
of the second kind so if you look at with
this kernel we have the Fredholm integral
equation of the second kind
And in place of k x t now we are writing the
summation i e equal to one to n a i x b i
t now this integration with respect to t I
can take this a i x common and since it is
find it’s sum we can interchange the summation
and integral sign so here we right a to b
b i t y of t d t now if you look at this quantity
is what a to b b i t y (t) d t this quantity
is going to be a constant value so if i denote
this constants
Say c i then we can write substituting this
c i as a to b b i t y (t) d t in this thing
then our solution y x can be written as f
of x plus lambda times this is c i c i into
t i x so it means that in this particular
case when kernel is given in separable form
if solution s given in this form y of x equal
to f of x plus lambda times i equal to one
to end c i e i x the only thing left here
is to find out these constants c one to c
n so how to find out these constants c one
to c n
So we can find out we can get the solution
so for this what we do we already know the
form of the solution you take this solution
defined in this and put it back into your
equation number 3
So here we have solution y of x is equal to
f of x plus lambda times summation i e equal
to one to n and c i a i x
Now your equation is what y of x is equal
to f of x plus lambda times 
summation a i x i e equal to one two n here
and this is integral a to b b i t y (t) d
t
Now y of t you can evaluate from this and
here we have c i is equal to a to b b i t
y (t) d t so what we do here we can write
it y(t) d t as f of t plus lambda times summation
i e equal to one to n c i or a i t we can
put the value of y t as this and simplify
it now here when you simplify it will be reducing
to this summation i equal to one to n e i
x here if you look at this we are assuming
a different as dummy variable say k so that
it should not
So k is from to n c k a k t so if we use this
we have this summation i equal to one to n
e i x into this quantity if you simplify you
will get this how we can get this let us look
at here I am writing y x here as this f of
x plus lambda times summation i e equal to
one to n c i a i x here equal to f of x plus
lambda times summation i e equal to one to
n a i x here a to b b i t here and in place
of y t I am writing f of t plus lambda times
since it is already inside
So I cannot use the same so let use k equal
to one to n c k a k t and d of t and when
you simplify this fx this will be cancelled
out here your lambda will also simply gone
and I can write it here that summation a i
x i e equal to one to n if you look at the
coefficient of this coefficient of this is
c i minus here i will get what I will get
this quantity so let me right it here a to
b b i t f of t here plus f of t dt here plus
I am taking here so 
minus a to b b i t and summation here
I will write it here summation k e equal to
1 to n b i t and a k t d of t equal to zero
anything we have missed a i x it’s ok b
i t a k t a k So we have this thing and I
can write it like this so this is given here
now this is written here now here we are using
the fact that this a i x are linearly independent
so if they are linearly independent means
the coefficient of a i x will be nothing but
zero so here this equation c i these are the
coefficient here
So this will equal to c i a to b b i t f t
plus lambda times summation c k a k t d t
is equal to zero here and if we simplify this
further by saying that a to b b i t f of (t)
dt is denoted as f of i and a to b b i t a
k (t) d t is equal to a i k then we can write
down the simpler form of equation number five
as this so here I can write it c i minus lambda
times k equal to one to n a i k c k equal
to f of i
So what I am trying to do here since just
I am summarizing here
Since a i x are all linearly independent so
this will b e simply zero so this will be
zero means I am just writing here equal to
zero for all i equal to one two up to n is
it ok and this I am writing as b i t this
I am writing as f i so let me erase this I
am writing this as f of i and this I am writing
b i t a k t as a i k so let me write it here
this is I am writing here a i k so I can simplify
this so here I can write it here c i minus
summation some lambda is missing here
I think lambda is there some lambda is there
so let me write it here lambda c k lambda
c k is missing here this is lambda c k here
so let me write it c i lambda c k and we have
a i k here and k is from one to h here is
equal to f of x so that is what is written
as equation number seven here is it ok so
now this is a simple algebraic equation which
we can solve by your knowledge of linear algebra
but here I just want to put one another way
to find out this c i
If you look at what we did here
Once we have a solution here we have put it
back to three but rather than putting it back
into three we can simply right it here look
at this the coefficient c i is given by c
i equal to a to b b i t y t so if you use
y t which is given by this I can write it
y t equal to f of t plus lambda times i equal
to one to n c i a i t if you put it back you
will get what let me write it here
These are your c i and I am writing a to b
b i t and in place of f t I am writing what
sorry y t I am writing f of t plus lambda
times summation k equal to one two and I am
writing k equal to one to n c k a k t and
this is d t and if you simplify you will get
what if you take it this here a to b b i t
f of t as f of y so this is f i plus lambda
times summation c k and b i t a k t we are
denoting as a i k so this is a i k k is from
one to n so here we get the same equation
So c i equal to f i plus lambda times k equal
to one to n c k a I k which is same as what
we have already obtained here so this is an
alternative way for obtained the equation
for this c i’s once we have c i whether
you obtain through this equation number three
or values from the coefficients so here what
we have seen here we can simplify the system
of linear equation into this part d lambda
c equal to f
So here your d lambda is i minus lambda a
where a matrix is given by a i j and c is
your c i and f is nothing but small f i basically
we are writing in this particular format so
we are writing here say i minus so here we
are writing here c i minus lambda summation
k equal to one to n c k a i k f of i for every
i equal to one to n so if I write it for one
so it is c one minus lambda I will write it
here c one a i c one k c one one let me write
it here
So here I will write it for so i equal to
one we have c one minus lambda c one a one
one minus lambda c two a one two and so on
and this minus lambda c n a one n equal to
f one and for i e equal to two it is c two
minus lambda c one a two one minus lambda
c two a two two and so on minus lambda c n
a to n equal to f one and in so on if we write
it i e equal to n it is c n minus lambda c
one a n one minus lambda c two a n two and
so on
Minus lambda c n a n n equal to f of n so
if you erase this you will get here first
we have one minus lambda a one one and here
we can write it minus lambda a one two and
minus lambda a one n and if you look at the
second one it is c one is here so minus lambda
a to one will be there and if you look at
this term it is one minus lambda a two two
and so on minus lambda a to n and in this
way we can write it here
If you look at the last one it is what minus
lambda a n one and it is what it is one minus
lambda a n n and this is c one to c n and
it is f one to f n which we are denoting as
d lambda c equal to f now here cases arises
when we discuss the coefficient matrix d lambda
and the forcing term this f now let us discuss
cases here so this is the equation 9 I am
representing here
Now this f which is nothing but this a to
basically f is your f i t
Now if the function f x is ideally equal to
zero in this two so what is two here two is
this so if I assume a homogenous Fredholm
integral equation then in this case this f
i if you look at equation number 6 a to b
b i t f of (t) dt equal to f of i and if f
of t is equal to zero all your f i is going
to be zero so in this case if this equation
is reduced to d lambda c equal to zero this
is happening when we are assuming that f of
x is ideally equal to zero
So when we take this f of x is equal to zero
or we can say that we are taking the homogenous
Fredholm integral equation then you can solve
our constant using this now here we have two
cases arise that if determent of b lambda
is zero and determent of d lambda is non zero
so in this particular case when we have determent
of d lambda is equal to zero then we have
a non trivial solution and in this case we
can say that we have a non trivial solution
And not only one solution we have infinite
number of solution here so in this particular
case when determent of d lambda is equal to
zero here we have infinite number of solutions
and if determent of d lambda is non zero then
we have a unique solution that is c equal
to zero so it means that all c i’s equal
to zero for i equal to one to n here and in
this case if solution y of x is given by one
the solution s always define as s
If you remember the solution is given by this
formula y of x equal to f of x plus lambda
times summation c i a i x i equal to y to
n so in particular this case when determent
of d lambda is non zero all those c i is equal
to zero we have already assumed that fx equal
to zero so in this particular case your y
of x is nothing but zero solution so this
is a zero solution for what this is zero solution
for homogenous Fredholm integral equation
So this is the solution for what this is solution
for this particular problem so here we have
y of x equal to f of x is simply zero here
a to b k of xt and y of (t) dt lambda here
and here k x t as a degenerate or a separable
kernel i equal to one to n a i x and b i t
so here when determent of d lambda is non
zero we have a trivial solution for this homogenous
Fredholm integral equation now for this particular
case when determent of d lambda is zero we
have a non trivial solution
It means that they are some c i’s which
are non zero and with the help of this your
solution you can write it here as let me write
it solution is basically what solution here
here fx is zero so we can write it yx plus
equal to lambda times summation i equal to
one to n and we have some c i zero so it means
that in this case when determent of d lambda
is equal to zero the solution s given by this
and we have infinite many solutions in this
case ok
So now let us move to next case next case
is that this is what we have discussed here
that if d lambda is equal to zero means determent
of d lambda is equal to zero then at least
one of the c’s can be assigned arbitrary
and the remaining c’s can be accordingly
determined or in this case we have infinite
number of solutions
And those values for this d lambda is equal
to zero d lambda equal to zero means the determent
of coefficient matrix is zero
Those values of lambda are known as eigen
values of this system so we call this as lambda
for which we have a solution here we say that
these lambdas are eigen values of homogenous
integral equation and the corresponding non
trivial solution we call this as corresponding
eigen functions so those values of lambda
for which d lambda is zero called as eigen
values and the corresponding non trivial solutions
are known as eigen functions
It may happen that a eigen value may have
more than one independent non trivial solutions
so in this case we may have corresponding
to one eigen values may have more than one
eigen functions available to us now let us
consider the second case the second case when
we don’t have this function f of x is not
ideally equal to zero when we don’t have
f x not ideally equal to zero but in this
case we may have two cases
One case is that if you look at f i is basically
what
F i is basically a to b b i t f t b i (t)
dt now it may happen that though your function
f of x is not ideally equal to zero but this
ft is orthogonal to this b i t it may happen
that this ft is orthogonal to this b i t in
this case also your f i is simply (28:23vanishing)
for all i equal to one to n so this may happen
when f t is orthogonal to your b one t to
b n t in this case also your problem reduce
to this d lambda c equal to zero right
So again as we are discussing in as a previous
case here also we have two case arise where
d lambda is zero and d lambda non zero right
and in this case your f x is non zero but
still because of this property orthogonal
d we have again the homogenous equation and
if we have d lambda equal to zero we have
more than one solution available so we can
say infinite number of solutions here and
solutions are given as y of x equal to f of
x plus
It is given by this particular formula this
thing
f x plus lambda times summation i equal to
one to n c i a i x right so d lambda equal
to zero means they are more than one solution
for c i’s so it means the infinite number
of solutions of c i available so we have infinite
number of solutions available here and when
we have d lambda is not equal to zero in this
case we have only unique solution that is
of c equal to zero right
And in this case the solution will be y of
x is nothing but f of x right so if you compare
these two cases when f x is equal to zero
and when f x is non zero but orthogonally
condition is there then here in this particular
case your solution is y x is equal to lambda
times summation c i a i x i equal to one to
n so in this particular case the solution
is not involving any function f of x but here
since f x is non zero solution is given in
this form
So here we have infinite number of solutions
and in this case when d lambda is non zero
you have y x equal to f of x here and here
we have only y x equal to zero now consider
one more case when f x is non zero and this
property is not true for all i equal to one
to n it means that it may happen that sum
of f i is zero but not all f i’s are zero
in this case 
it is d lambda c equal to your f but here
your f is not a zero factor now
Now again we have two condition d lambda is
equal to zero and d lambda non equal to zero
now when d lambda is not equal to zero then
here we have the unique solution that is c
equal to d lambda inverse f so we have a unique
solution and solution is given by again y
of x is equal to f of x plus lambda times
summation i equal to one to n c i a i x so
c i you can find out using this and you can
put it here and you have a solution
Now when we have d lambda equal to zero then
here we have coefficient matrix which is a
similar coefficient matrix and here f is a
non zero factor it may happen that we don’t
have any solution at all here your knowledge
of solving linear algebraic equation is very
very important if you know how to solve a
x equal to b you can solve this thing so it
may happen that when d lambda is equal to
zero it may happen that there is no solution
at all for this
That you check from the rank condition here
so if rank of d lambda and rank of d lambda
f are same then we have a solution and in
that case we have infinite number of solution
and if rank of d lambda and rank of d lambda
f is not same then we have a no solution so
here we have two conditions no solution and
infinite number of solution 
that depend on rank condition that here no
solution means rank of d lambda is same as
rank of
I hope this you have already seen d lambda
f and if it is not same then no solution if
it is same then we have infinite number of
solution so here we have condition rank of
d lambda is equal to rank of d lambda f so
in this particular case we have infinite number
of solution here we have no solution so now
we have discussed all the cases I hope we
have considered all the cases ok so in next
lecture we will discuss the example based
on this theory
Is it ok thank you for being with us
