Hello friends, welcome to this lecture. In
this lecture we continue our study of boundary
value problems. And in this lecture we will
discuss the very special kind of boundary
value problem that is Sturm-Liouville boundary
value problem. So let us first define what
is a Sturm-Liouville boundary value problem.
So first we discuss the differential equation
of the form py dash dash+qy+lambda ry = 0,
here. So here this is a by looking we can
say that this is the self adjoint differential
equation, here p dash q and r are real- valued
continuous functions on close interval A and
B and lambda is a real parameter. And also
let us assume that this p x is never 0 in
this interval, so without loss let us assume
that P(x) is > 0 and also we assume that the
coefficient of y function that r(x) is also
> 0.
So this is known as Sturm-Liouville equation.
Sturm-Liouville equation without putting this
condition that p x and r x are non-negative.
So here we simply consider this equation given
is equation number 12, we say that it a self
adjoint equation and given in this form and
we call this as Sturm-Liouville equation.
And with the condition that p dash q and r
are the real-valued continuous functions.
And here we consider the boundary conditions,
so Sturm-Liouville equation along with boundary
condition then it is known as Sturm-Liouville
boundary value problem. So boundary conditions
are given either in this form or this form.
So here the first set of boundary condition
is this that m1y of A + m2y dash A=0 and m3y
of A + m4y dash A= 0 here. So here boundary
condition is given in 13a and 13b.
So here if you look at the one condition is
given at the point x=A; t=A and the other
boundary condition given in the at the point
of B, so here we have m1y of A + m2y dash
A=0 and m3y of B+ m4y dash B=0, so this kind
of boundary conditions are known as separated
boundary conditions because they are separately
defined at point A and B. And there is one
more kind of boundary condition that is periodic
boundary condition that is y of A= y of B
and y dash A=y dash B.
And if you are considering one more condition
that is p of A= p of B. So we call these kind
of conditions has periodic boundary conditions.
So initial whatever be the value at the initial
point that is value given at the end point,
so it means that y of A = y of B;’ y dash
A=y dash B and so on. So and here we are assuming
that in 13 the variable m1, m2 must non-zero
similarly one of the coefficient that is m3,
m4 are non-zero constant.
So here we assume that m1 square+m2 square
is non-zero, similarly m3 square+m4 square
is non-zero. So at least we are saying that
these are not trivial boundary conditions.
So a boundary value problem consisting of
equation number 12 that is Sturm-Liouville
equation, along with boundary condition whether
it is 13a or this 14, this, we say that strum
is called regular Sturm-Liouville boundary
value problem. Here regular means what? First
of all, we define what is Sturm-Liouville
boundary value problem for defining Sturm-Liouville
boundary value problem we consider Sturm-Liouville
equation that is SL equation.
And one of the boundary conditions either
it is 13a or 14 then we call this as Sturm-Liouville
boundary value problem. And then we define
a regular Sturm-Liouville boundary value problem,
it means that our problem is regular. In the
sense that, your point A, B these are finite
end point that none of A and B are infinite,
and the coefficient of highest order derivative
that is y double dash that is p here is non-zero
in this interval.
If it is happening then we call this SL boundary
value problem as regular Sturm-Liouville boundary
value problem, okay. So here we say that we
already know that, since it is a homogenous
equation so trivial solution is always exists.
So it means that ideally 0 function on A,
B is always a solution of Sturm-Liouville
problem and we see that this solution is a
trivial solution of this boundary value problem.
But, we are more interested in finding the
non-trivial solution. So we say that for a
value of this parameter lambda let us say
that y lambda is a non-trivial solution of
12 with 13a, 13b or 12 with 14. Then this
parameter the value of this parameter for
which we have a non-trivial solution the value
is known as eigenvalue and the corresponding
solution is known as eigenfunction.
So our idea of considering the Sturm-Liouville
boundary value problem is to find out the
parameter lambda for which this has a non-trivial
solutions and we see that this has a very,
very useful application and in fact we say
that the application is one very important
application is that, that this system has
a complete orthonormal system of eigenfunctions
and we use these eigenfunctions in expanding
a certain class of function.
That we see as an application of a Sturm-Liouville
boundary value problem. And you can see that
some very important boundary value Sturm-Liouville
boundary value problems are example of Legendre
polynomial and Bessel polynomials and we say
we have already seen that how a given function
can be written as expansion of Legendre polynomial
or Bessel polynomials. So here we see the
general theory behind this.
So here idea is that in throughout this lecture
we always try to find out the value of the
parameter for which we have a non-trivial
solution. And this set of all eigenvalues
of a regular problem is called its spectrums.
So set of eigenvalues is known as spectrum.
So now let us consider one example of finding
all the eigenvalues and eigenfunctions of
the boundary value problem, so if you look
at y double dash + lambda y=0. So first let
us see that this equation is given in terms
of self adjoint form. So here coefficient
of y double dash is 1 and coefficient of y
dash is simply 0, so we can say that this
equation is given in self adjoint form.
Now, we have a boundary condition, boundary
condition is given as y0=0 and y pi=0, so
these boundary conditions are separated boundary
conditions, so it means that this system of
boundary value problem is a regular boundary,
regular Strum-Liouville boundary value problem
because here endpoints are also finite and
coefficient of y double dash is non-zero.
So we say that this system is a regular Strum-Liouville
boundary value problem.
And we want to find out the parameter lambda
for which we have a non-trivial solution.
For which, let us consider several cases.
So since lambda is a real numbers so it can
take either 0 value, positive value or negative
value, so we will consider one by one. So
first let us consider the simplest case that
is lambda=0 and in this case equation reduced
by y double dash=0. And we already know that
the solution is simply a linear solution so
it means y of x=A of t+B t A t+B and we can
find out the coefficient A and B, sorry it
is y of t not y.
So you can find out the coefficient A and
B by this y of 0=0 and y of pi=0 and you can
simply say that since y0=0 this implies at
b=0. Now y pi=0 means you can say that we
have that A=0. So in case when lambda=0 then
we have only 0 solution that y t=0 is the
only solution. And we say that we do not have
a non-trivial solution, so this is not very
useful to us and we simply say that lambda=0
is not an eigenvalue because eigenvalue we
say when we have a non-trivial solution corresponding
to the value of the parameter.
So we say that lambda=0 is not an eignvalue
of equation number 15. So now let us consider
the next case that is lambda < 0. So when
lambda is < 0 we can write it lambda as –Mu
square where Mu is some real number. And we
say that solving this we have this equation
y double dash –Mu square y=0 and we already
know the solution of this, solution of this
is basically what, here we write it—
y double dash – Mu square y=0, so we can
write it solution of y t = e to the power
Mu t, let me write it A+B e to the power –Mu
of t. And we need to find out the coefficient
A and B for that we let utilize this condition
y 0=0 and y of pi=0 when you apply y 0=0 then
it is A+B=0 and when you apply Y pi=0 that
is e to the power pi t + B e to the power
–pi of t=0. And when you simplify this that
is 1 1 e to the power pi t and e to the power
–pi t and here we have A B=0 0.
And you can simply say that the determined
of this coefficient metrics is non-zero so
it means that this is invertible and you can
say that your solution is only that A=0=B.
So it means that here we have only A trivial
solution because here A and B are 0 so we
have only A trivial solution. So we say that
we does not have any non-trivial solution
so lambda < 0 is also not the eigenvalue.
So it means that any lambda for which, any
non-negative value of lambda is also not an
eigenvalue of 15.
So now let us move to next condition that
is lambda > 0. So for that let us assume that
lambda=Mu square where Mu is any real number
and hence we can write it our equation as
y double dash + Mu square y=0 and we already
know that solution is y t = C1 cos of Mu t+C2
sin of Mu t, and we need to find out this
C1 and C2 for that y0=0=y of pi. When you
put y of 0=0 so the 0=, here we have C1 because,
so C1 is coming out to be 0, now y pi=0.
When we put y pi=0 we have 0=C2 sin of Mu
pi, so it means that we have C2 sin Mu pi.
Now we already know that if C2 is 0 then we
have only a trivial solution which is not
useful to us, so let us find out the values
of Mu say that this C2 sin Mu pi has to be
so we say that if Mu of pi is n pi where n
is 0, 1, 2, 3 and so on then we simply say
that we have a non-trivial, in that case we
can choose any value of C2.
So here we can say that when lambda is > 0
then we can write lambda=Mu square as we have
pointed out, solution is given as y t=C naught
cos of Mu t+C1 sin of Mu t and when you use
y0=0 you know that C naught = 0 and y pi=0
and pi z C1 sin Mu pi=0. And we have a non-trivial
solution only when Mu pi= n pi where n is
any integer value. So it means that calling
that-- we say that for each n= each n belonging
to z we have solution that is sin of n t,
right.
And in this case we say that our lambda n
which is Mu square that is any square is the
eigenvalue so any square is the eigenvalue
so and the corresponding solution is sin of
n t. So you’re your eigenvalues are what
eigenvalues are every integer value is an
eigenvalue and this corresponding eigenfunction
is given as sin of n t. So here we are able
to find out eigenvalues and eigenfunction
of a given Sturm-Liouville boundary value
problem that is this here.
So now proceed further.
And we have a-- this very, very important
result which says that assume that A and B
are finite real number, so you have a finite
endpoint. And the functions qt and rt are
real value continuous function on this close
interval A, B and that m1, m2, m3 and m4 that
is the coefficient present in a boundary conditions
are real numbers then the Sturm-Liouville
problem 12 with 13a, 13b or 12 with 14.
Now 12 is what, Sturm-Liouville equation,
and 13a, 13b are separated boundary condition
and 14 is the periodic boundary condition.
So we say that if A, B and C satisfy then
we have countably many eigenvalues with no
finite limit point, corresponding to each
eigenvalue we have eigenfunctions. So this
is the existence theorem of eigenfunction.
Here I am not going to prove this, we assume
the statement and the result of this theorem
as a result which gives us license to work
with eigenvalues and eigenfunctions.
So without proving without providing the proof
of this we are using this theorem as an existing
theorem for eigenvalues and eigenfunctions.
And we say that eigenvalues are countably
many and corresponding to each eigenvalues
we have eigenfunctions. Here we have a eigenvalues
are like discrete eigenvalues and we say that
discrete eigenvalues means these are not continuous
spectrum it is only discrete one, okay.
So all these we have utilized so it means
that in regular Strum-Liouville boundary value
problem we have this theorem to provide the
existence of eigenvalues and eigenfunctions.
Now once we have existence of eigenvalues
and eigenfunction then we start working with
these eigenvalues and eigenfunctions.
So next very important result is in the direction
of orthogonal eigenfunctions. So before that
what do you mean by orthogonal eigenfunctions
let us define it. So two distinct functions
x and y, defined and continuous on this close
interval A, B are said to be orthogonal with
respect to a continuous weight function r
t if, A to B r s, x s, y s ds=0. So here when
this condition hold then we call this as that
x and y are orthogonal to each other with
respect to the weight function r of s.
And this is very, very important property
which is preserved by if you look at Legendre
polynomials, Bessel polynomial and many more
special function which is nothing but a bi-product
of the Sturm-Liouville boundary value problem.
So this is very, very important property,
and it says that this Regular Strum-Liouville
boundary value problem consist a complete
orthogonal system of eigenfunctions. So let
us that we are going to discuss in next class.
Now let us discuss a next theorem.
So let all the assumptions a and b of Theorem
15, Theorem 15 is existence of eigenfunctions
hold and for the parameter lambda and Mu where
lambda is not equal to Mu, let x and y be
the corresponding solution of 12 such that
this value that is p Wronskian of x, y given
at B and A=0, basically it is what it is p
w, x y given at B – or you can = p w, x
y given at A are same. So this quantity is
same. Then we say that A to B r of x, x of
y, y of x ds = 0.
So we say that if this condition hold what
are conditions, condition is that A and B
are finite real number and the function p
dash t, q t and r t are real valued continuous
functions then the Strum-Liouville boundary
value problem has orthogonal solution corresponding
to non-distinct eigenvalues. So it means that
even lambda and Mu are not equal then corresponding
to lambda let us say x is the solution of
the Strum-Liouville boundary value problem
and corresponding to Mu we have y has solution
of the boundary value problem.
Then x and y are orthogonal to each other
with respect to the function r of s. So let
us prove this and this is true when we have
the following condition that is this.
So let us proof this. So from the hypothesis
of the theorem we know that corresponding
to lambda x is the solution of this Sturm-Liouville
equation. So px dash dash + qx + lambda rx=0
and py dash whole dash+qy+Mu ry=0, so that
is what is given here that x is the solution
of this and y is the solution of this, so
it satisfy this. Now simplify this by multiplying
the first equation by y and multiply the second
equation by x and just subtract it.
So here when you multiply what you will get,
this Sturm will be canceled out qx y-qy x
it is gone. So here in, let me write it here,
so it is what p x dash dash y+q x y+lambda
r x y=0 and here we have p y dash x+q y+Mu
rxy=0, when you subtract it this sample gone
and we have lambda–Mu rxy, so we have lambda–Mu
rxy=0 p y dash whole dash x-px dash dash y,
so we have this thing.
Now, we can simplify and we can write it lambda
– Mu rxy=dy/dt of py of x-px dash y. Here
we simply add this term py dash x dash and
we can write it like this, in fact you can
easily check that if you expand it you will
get this only. So here we have done this p
y dash dash x+py dash x dash - py dash x dash
– px dash whole dash y, and if you simplify
you have this thing, that d/dt of py dash
x–px dash y.
So then we integrate this to with respect
to t from A to B. So when you integrate this
last equation 17 we have lambda – Mu A to
B r of s, x of s - t s=, now here this derivative
will go out and we have p y dash x–p x dash
y evaluated at B-A. So now look at this equation
is already given. What is given here that
p Wronskain of x y at A to B = 0, means what
p-- now Wronskain x y is x y dash – x dash
y given at the = 0, so this already given
which is written here.
p x dash y B at A. So here we can say that
quantity is going to be 0 because that is
given as assumption here. This is the condition
we have given. So it means that this right
hand side is simply gone so we have lambda
– Mu*A to B r s, x s, d s = 0. Now lambda
is not equal to Mu that we have already assumed,
so it means that from the assumption we have
that this quantity has to be 0.
That is what we wanted to prove that the solution
corresponding to different distinct eigenvalues
are orthogonal to each other that is what
we wanted to prove.
Now next we want to show that, that we have
already proved that if certain condition hold
that is p of Wronskain of x y add B=p w xy
at A if these two are equal then distinct
eigenfunction corresponding to distinct eigenvalues
are orthogonal to each other. Now see that
how this boundary condition which we have
assumed that is separated boundary condition
and periodic boundary condition are helpful
to find out that this condition is trivially
hold.
So for that let us consider this following
theorem. Let the hypothesis of Theorem 15
that is the existence of eigenfunction theorem
be satisfied and let xm and xn be two eigenfunctions
of the boundary value problem corresponding
to two distinct values lambda m and lambda
n. Then this boundary condition holds true.
Because if you look at, we have assumed that
if this boundary condition hold then the corresponding
eigenfunction that is xm and xn are orthogonal
to each other with weight function r of x.
Now you want to show that if we have boundary
condition that is 13a 13b is true then this
condition is automatically true. Now here,
in addition to this we say that if p A=0 then
this condition holds without use of 13b, right
13a. And if p B=0 then this equation holds
with the 13b being deleted. So it means that
if p A=0 then we have to use only one condition,
if p B=0 then we can use only one condition.
So without loss of generality let us assume
that, that none of this p A and P B is 0.
So let us assume that p A is not equal to
0 and p B not equal to 0. In fact let me write
it here this is what this is p B Wronskian
of x of xn given at B–p of A Wronskian of
xn at A=0. Now if p B is 0 then I need not
to use this only this is sufficient that is
what it is written here. Then if p B=0 then
we have to use only one that is 13a.
And if p A=0 then we have to use only one
condition that is 13b and the boundary condition
given and the point t = B. So let us assume
that, that none of this p A and p B is 0.
So from this theorem we have, that since xn
and xm both are solution of this so let us
write it m1 that they must satisfy the boundary
condition that is m1 xn A+m2 xn dash A=0,
similarly xn will also satisfy the boundary
condition that is m1 xn A+m2 xm dash A=0.
So here let us assume that this m1 is not
equal to 0 because if m1is =0 then this is
quite easy, we see what is going to be easy
because if m1 is not equal to 0 then we can
simplify these two equation by eliminating
this m2, so here you can eliminate m2 from
these two equations and we can write our equation
as m1 xn A xm dash A–xm A xn dash A=0. So
here we assuming that m1 is not equal to 0.
If m1 is equal to 0 then this is very easily
we can obtain like this but coefficient is
going to be m2. Now we have this, so since
m1 is not equal to 0 so we can cancel it out
and we have this thing.
That xn xm dash A - xm A xn dash A=0, so this
we have obtained from 13a that is the boundary
condition given at the point t=A. Similarly,
if we assume that use 13b is let us assume
that m3 is not equal to 0 then in a similar
manner we can prove that, that xn B xm dash
B - xn B xm dash B=0. And we can easily see
that if p A=0 then this 19 is not useful.
And if p B=0 then this is not useful.
And we can say that from equation number 8
let us simplify this p w xn xm given at B
at A=p B xn B xm dash B –xn B xm dash B.
Now since p A=0 we are not looking at the
condition given at xt=A, so we have this.
Now if you look at the in a bracket whatever
is here it is trivially 0 because we have
already proved in equation number 12. So it
means that boundary condition 13b implies
that 20th is true.
Now 20 is true means Wronskian this pw xn,
xm given at the B and A=0 if p A=0. Similarly,
we can handle the case and p B=0. So it means
that if p A and p B both are non-zero then
we can simply say that this is what this is
written as p B and this is xn B, xn B xm dash
B – xn B xm dash B – p of A xn A xm dash
A – xn A xm dash A, so we have from 19th
this is 0, this is 20th this is 0, so it means
that quantity is going to be true.
So we have seen that the condition this condition
that is p w xn xm at A and B=0 is true when
we assume the condition that 13a and 13b that
is the separated boundary condition if you
use then this is automatically true.
So now we look at the other case and consider
that hypothesis of Theorem 15 we satisfied
and let xm and xn are eigenfunctions of the
boundary value problem 12 and 14 that is periodic
boundary conditions are true corresponding
to distinct lambda m and lambda n then xm
and xn are orthogonal with respect to the
weight function r t.
In fact, only thing we had to show is this
quantity this is going to be 0 here. So let
us look at here we since p w xm, xn B-- A
is can be expanded in this form p B * B Wronskian
of xm, xn that is xn B xm B-xm dash B xm B-p
of A xm A – xm dash A+xn dash A+xm A. Now
here we simply say that if we use periodic
boundary condition then we can make this quantity
0. So here we use xm B=xn A. xm dash B = xn
dash A; xn dash B = xm dash A; xn B = xn m.
And we have also assume that p of B= p of
A. So we can say that here this periodic periodicity
if you apply then this quantity is =0, so
it means that if this quantity is 0 we have
already proved that if this condition is true
then A to B r s, xn s, xm s d of s = 0 here.
So in case of periodic boundary condition
that is p B=p A; x A=x B and x dash A=x dash
B, this quantity is 0 and hence your xn and
xm are orthogonal to each other with respect
to the weight function r of s.
And previous theorem says that if we have
the separated boundary condition then this
is automatically true, here we need not to
assume that p A = p B. But when we consider
the case of periodic boundary condition then
we have to consider the function p x also.
So here we have discussed some important properties
of eigenfunctions, what we have and discussed
is that we are considering the regular Strum-Liouville
boundary value problem.
And we have seen that, that eigenfunctions
existence of eigenfunctions that is theorem
15 we have discussed and not only this, we
have discussed that when the case of separated
boundary condition or periodic boundary condition
the regular Strum-Liouville boundary value
problem has orthogonal set of eigenfunction
corresponding to distinct eigenvalues.
So and in next class we will some more properties
of eigenfunction and eigenvalues and the use
of eigenfunctions in many other field. So
with this I am just summarizing my lecture
and we continue in next lecture. Thank you
very much for listening, thank you.
