Welcome back in the last few videos we have
seen one example of regular sturm louisville
system to find the eigenvalues and eigen functions
and rest examples for periodic and singular
sturm louisville systems we have seen Today
we will just have we will have 2 more examples
of different way of finding eigenvalues and
eigen functions for the regular sturm louisville
system So to start with the 1st example which
i explained in the last video So we can see
this this is the example so for which you
can find eigenvalues are eigen functions so
little different from the work we have done
for the 1st example of regular sturm louisville
system
So you need to take the differential equation
plus lambda y equal to 0 so this you want
to put it in this sturm louisville system
so self a joint form or hermitian form or
skew symmetric form so whatever you say So
x is between 0 to 3 this is what is the domain
You need to put this together as derivative
of something so to do this this is just a
linear equation of time right So if you think
of y – is a dependent variable you have
dy – by dx minus2 dy by dx okay So that
dy – minus 2y – So for y – is a linear
equation so you know that integrating factor
is e power integral p p is here minus2 dx
So you have e power minus 2x that is the integrating
factor for these 2 terms So if you multiply
this equation e power minus 2x so e power
minus 2xy double dash minus 2 into e power
minus 2xy – plus lambda into e power minus
2xy equal to 0 so that x is between 2 pie
Now i can put this with multiplication of
the integrating factor we can combine these
2 as derivative of e power minus 2xy – or
dy by dx Okay 1 you can see that e power minus
2x y double dash and if you differentiate
this you get e power minus 2 into the e power
minus2 xy – So plus q0 and you have you
can put this as outside minus and this will
be lambda into e power minus 2xy okay
So this is actually your ly ly has a definition
so x is between 0 to pie and l is minus ddx
Now p is minus e power minus 2x okay And q
is 0 so 0 into 0 simply q is 0 so we do not
write this So this is what so once you identify
this operator we can write the dot product
of 2 functions Here the definition is from
this domain which is 0 to pie and w so w is
you have a w here so e power minus 2x so so
you can just divide with this okay so you
can write 1 by e power so you can write bring
this e power minus 2x this side that is going
to be plus okay This is going to be plus so
you can remove this so where l is 1 over e
power minus 2x
So this is your w so w is this into fx gx
bar dx this is a dot product you have So this
is what is your sturm louisville equation
ly equal to lambda as the eigenvalue problem
Okay So you need to find the eigenvalues and
eigen vectors Now consider so consider now
l is the self adjoint from so you have hermitian
form implied lambda is real Lambda is real
since l is in hermitian form or skew symmetric
form So we can now consider so lambda is real
means it can be positive negative and all
such thing You 1st consider the general solution
of the equation given equation
So that is y double dash minus 2y – plus
lambda equal to lambda y equal to 0 Okay so
this form what are the solutions this is the
linear equation with constant coefficients
you can look for solutions in this form e
power kx If you look for this form you put
it you get k square minus 2k plus lambda equal
to 0 So that will give me whatever my k values
through plus or minus square root of b square
minus4 lambda divided by 2 So this will give
me 1 plus or minus square root of 1 minus
lambda So these are your roots so implies
you have this general solution general solution
is yx which is the second-order equation so
you can get the general solution as some c1
times
When you see here this is c1 times e power
1plus square root over minus lambda into x
plus c2 e power one minus square root of 1
minus lambda into x So this is your general
solution Now you know that lambda is real
we just start with because see it involves
the general solution involves 1 minus lambda
so let us consider 1st lambda is equal to
1 okay So you can also think if it is not
there so you can you can also look at lambda
positive lambda negative and lambda equal
to 0 Because it has at 1 minus in the solution
you have square root of 1 minus lambda that
suggests you work with lambda equal to 1 So
if you put lambda equal to1 that is your general
solution so what happens to your roots
With lambda equal to1 what you get is k k
minus 2k plus1 equal to 0 So you have k minus1
whole square equal to 0 that will give me
k is 1 and 1 So you have repeated roots so
the general solution is 
general solution is e power c1 e power kx
k is 1 so you have e power x plus c2 x into
e power x okay So this is what you have now
you apply boundary conditions What you have
is starting with the 1st boundary condition
y0 is 0 if you apply here and so y0 is 0 so
you have c1 so c1 plus c2 into 0 So that is
0 so this is 0 so you get c2 is 0 c1 will
be 0
So this implies the general solution becomes
c2 x e power x Now you can apply other boundary
conditions So apply the boundary other boundary
conditions that is we have y pie equal to
0 So if you apply this you get c2 by e power
by has to be 0 this is possible only if c2
c2 is 0 right So you have c2 is also 0 this
is positive since this is nonzero so c2 has
to be 0 So general solution finally that implies
general solution or the solution satisfying
the boundary condition becomes why x is identically
0 So that means lambda is equal to 1 is not
an eigenvalue Okay So we can look at lambda
greater than 1 and the lambda less than 1
You have 3 cases that we can look at lambda
greater than we look at lambda greater than
1 then in that case this is your general solution
and this will be i okay So when lambda is
greater than 1 general solution of the equation
yx equal to so e power x is common so you
can look at this e power one x he power one
x is common so you can take this out and you
have c1 and what you get is when lambda is
greater than 1 1 minus lambda is negative
so you have lambda minus1 i times square root
of lambda minus1 lambda is greater than 1
So you have c1 e power minus so you can rewrite
so you have that involves e power i square
root of lambda minus1 x here you power minus
i square root of lambda minus1 x
So that you can write also as a linear combination
of cosine and sines So we will write as cos
lambda minus1 x under root of whose square
root of lambda minus1 into x okay and then
plus c2 sin square root of lambda minus1 into
x x is outside okay So this is your general
solution now you apply the boundary conditions
Y00 will give me what happens to 0 e power
0 is 1 c1 that is c1 plus c2 when you put
x equal to 0 that is 0 so this is what you
get c1 this is y0 so this is equal to 0 So
you get again c1 is 0 so the yx becomes e
power x c2 sin square root of lambda minus1
into x
Okay So now we apply another boundary condition
y pie is 0 that will give me e power pie c2
e power pie sin lambda minus1 into pie equal
to 0 So this is nonzero this cannot be 0 and
this can be 0 for some lambda values for lambda
greater than 1 So we have to see what are
these values okay So for such lambda values
c2 can be arbitrary So you can take this lambda
minus1 into pie equal to 0 will give me square
root of lambda minus1 pie equal to n pie And
you see that this quantity lambda is greater
than 1 so this this should be running from
uhh running from 1 2 3 onwards
Reason is n equal to 0 if i take an equal
to 0 that will become 0 that is lambda equal
to1 which already seen that is not an eigenvalue
If i take n negative then lambda minus1 is
actually it cannot be negative so it has to
be positive So this is what you are forced
so this sin values positive value so square
root so square root of this number should
be positive So right so so we have this So
this implies you have lambda minus1 under
root is n this will give me lambda as n square
plus1 n is running from 1 2 onwards So these
are your eigenvalues lambda n equal to n square
plus1 eigenvalues for which you have nonzero
solution So you can you have c2 can be arbitrary
you can take it as 1 e power x for these lambda
values
If you put it so eigen functions will be you
call them vn of x which is e power x sin root
lambda minus1 is n so you have sin nx So for
both n is running from 1 2 3 and so on Even
if we put n equal to 0 actually becomes you
see that vn is 0 So it is not an eigenvalue
corresponds to lambda 0 equal to1 So these
are your eigenvalues and eigen functions when
you consider lambda greater than 1 So you
have another case that is lambda less than
1 if you consider this case the general solution
we can rewrite So yx is eta were x is common
so you have c1 e power and lambda is negative
when lambda is less than 1 this will be positive
okay so this will be positive So you can write
as it is
So you have simply write 1 minus lambda x
plus c2 e power minus square root 1 minus
lambda x okay So now you apply the boundary
conditions y0 is 0 implies c1 plus c2 equal
to 0 Okay So that will give me c1 equal to
minus c2 so implies y x equal to e power x
c1 you can take it out e power x is common
what you have is e power square root of 1
minus lambda x and you have c2 i am replacing
with minus c1 so you have minus e power minus
square root of 1 minus lambda x Okay So this
is what you have so this is your general solution
have to plan the boundary condition
Now you apply ghee at the boundary condition
y pie is 0 you get c1 e power pie the power
square root of 1 minus lambda pie minus e
power minus square root of 1 minus lambda
pie So this quantity has to be 0 And you see
that this is e power something fix lambda
less than one this is infinite number So e
power let us call this c c pie minus e power
minus c pie this will ever be 0 So this cannot
be 0 and this cannot be 0 that means c1 has
to be 0 So that implies yx is identically
0 That implies lambda less than one is not
an eigenvalue Because you do not have nonzero
solutions for the system
So what you have is what you have found this
only these are your eigenvalues and these
are all eigen functions So any piecewise function
any piecewise continuous function piecewise
continuous function f x which is the domain
between 0 to pie i can write can be written
as 
fourier series fourier series is fx is from
n is from 1 to infinity you have some cn and
what are your eigen functions e power x sin
nx e power x sin nx okay Where cns is simply
we integrate so take the dot product where
cns are how do we find my cns you mult you
just take the dot product with vns e power
x sin mx equal to so when you multiply this
e power x sin mx only cm will be contributing
m equal to n only will contribute
So you have e power x dot product sin nx mx
e power x sin mx okay That is cm is integral
0 to pie you have e power minus 2x is wet
function e power minus 2x fx e power x sin
mx dx So this e power x and e power minus
2x will go so you have e power minus x this
divided by now this one integral 0 to pie
e power minus 2x E power x sin mx whole square
so when you will square this is what you get
So this will go so what you get is this this
quantity is simply because integral 0 to pie
sin square mx dx is You can write 1 minus
cos 2 mx divided by 2 from 0 to pie that value
is pie by 2
Just the half that you put the limits 0 to
pie so you have pie by 2 so you have 2 by
pie cm is 2 by pie 0 to pie 0 to pie e power
minus x fx sin mx dx So that is what that
is what is your cm So these are your fourier
coefficients fourier transforms Fourier transform
with this with respect to these eigen functions
this is your fourier series fourier series
okay So because something 2 some simple some
other example not from in terms of sines and
cosines so you can say that is generalised
fourier series okay You can say generalised
fourier series and generalised fourier transform
This for this particular example this particular
fourier type of transform and series okay
And this way for a piecewise continuous function
fx this convergence of this series is actually
point wise You fix x as the series the series
of numbers this converges to a function value
at the point Okay So look at some other examples
of regular sturm louisville system in which
case so we may not be able to explicitly your
eigenvalues Okay Let us see that example
One more example with that we will wind up
sturm louisville theory So you can have so
what you consider is now find eigenvalues
and eigen functions for the regular sturm
louisville system regular sturm louisville
system that is y double dash plus lambda y
equal to 0 lambda is a parameter and you have
the domain is between 0 to 1 x is between
0 to 1 and you have y0 is 0 and y1 minus y
– 1 equal to 0 So because it is a regular
sturm louisville system you can give this
is simpler one but this is a combination of
its derivative So we have chosen combination
of y and its derivative at 1 this is equal
to 0
So if you have like this always when you take
the combination you may not be getting the
solution eigenvalues explicitly we will see
how it is So we write this as y dy by dx ddx
minus equal to lambda y this is my ly ly okay
so x is between 0 to 1 So it is already in
the self adjoint form So l is minus ddx of
ddx p is 1 1 into ddx and w is 1 q is 0 So
dot product of i can define dot product here
this system is 0 to 1 w is 1 so you have simply
fx gx bar dx So what happens now you find
the now it is in self adjoint form so lambda
is real since l is self adjoint l is self
adjoint or hermitian or you can say skew symmetric
Now you can look at all the different cases
So lambda is either positive lambda equal
to 0 or lambda less than 0 So we look at all
these cases okay so start with lambda positive
if you take the 1 if lambda is positive so
we have y double dash plus lambda y equal
to 0 So lambda positive means lambda equal
to mu square mu is positive as usual we can
think of like this we can put lambda equal
to mu square So we will get general solution
general solution y x is c1 cos mu x plus c2
sin mu x so this is the general solution Okay
between x is between 0 to 1
Now when we apply the boundary condition like
earlier we have so you have y0 is 0 will give
me c1 is 0 c1 is 0 that is 1 cos 0 is one
plus c2 into 0 that is 0 So i have c1 which
is equal to 0 So we have c1 is 0 So general
solution becomes now c2 sin mu x Now you apply
other boundary conditions that is y at 1 minus
y – at 1 equal to 0 If you apply the boundary
condition here for this you have c2 sin mu
minus c2 mu cos mu into 1 So that this this
has to be 0 so this is nothing but sin mu
minus mu cos mu has to be 0 So i just have
to see for those new positive values for this
point is 0
Is no such value satisfying this equation
that means c2 has to be 0 that means all new
positive are not an eigen values But we have
to see this one this is called dispersion
relation so sin mu minus mu cos mu whether
it has any positive mu solutions So this is
this is actually tan mu equal to mu Right
We can just divide taking that cos mu 
so we can rewrite this tan mu equal to mu
So sin mu divided by cos mu that means cos
mu should not be 0 right When cos mu equal
to 0 that is you know that these are pie by
2 3 pie by 2 and so on so these are the values
we know that this quantity is nonzero Okay
So this except except that this has to be
0 so cos mu should not be 0 when you do this
one when you divide Suppose it is 0 okay suppose
if it is 0 that is true only at these values
But these values you can easily see that this
quantity is nonzero So these mu pie by 2 3
pie by 2 this quantity is nonzero So you can
think of other values you can divide it and
you can now find the roots of tan mu equal
to mu Okay
So this we do graphically this is like uhh
let us say mu is here this is my tan x equal
to x okay this is your y Y equal to mu x y
equal to x is something like this line And
you have tan mu that is tan x y equal to tan
x so tan 0 is 0 tan pie by 2 is only you want
positive side it tan pie by 2 tan pie tan
3 pie by 2 and 2 pie and so on So for this
you have you have things like this when you
say pie by 2 it is going to infinity so you
have something like going to infinity at pie
by 2 Other things uhh you will have tan what
is the other value so that is tan pie and
what is tan pie sin pie by cos pie so that
is 0
So it should be 0 here and here at pie by
2 the negative side it should go to it should
nicely go to eventually asymptotically this
should go to minus infinity At 3 pie by 2
it should go to asymptotically minus infinity
Again at 2 pie by 2 and 2 pie plus pie by
2 is 5 pie by 2 And again here you have something
like this curve Okay this is at 2 pie so like
this you get So you see that this curve and
this car is touching at these points okay
When this is 0 so we do not consider so this
is your because mu is positive mu is positive
so you have to worry about only this 1 this
is your mu 1 this is your mu 2 and like this
and so on
You can denote them like that and say then
let mu 1 mu 2 and so on be the roots of you
have roots that is clear from this graph tan
mu equal to mu Then lambda is mu i square
y is from 1 2 3 and so on that is your lambda
i these are your eigenvalues for which c2
is arbitrary Once c2 is arbitrary so you have
yx is sin mu x sin mu x those are your vi
of x these are eigen functions Sin sin mu
i mu ix So these are eigenvalues then these
are eigenvalues So these are this and this
are eigenvalues and eigen functions respectively
This corresponds to eigenvalue and this corresponds
to eigen functions
So you do not know explicitly exactly what
are your mu i’s some values okay Wherever
so numerically you can find out all these
values mu 1 mu 2 3 and so on there will positive
okay So i form certain eigenvalues are eigen
functions corresponding to the case mu positive
that is lambda positive Now look at the lambda
equal to 0 case
Now you know that general solution of the
equation is y double dash is 0 is equation
1 lambda equal to 0 So the general solution
is c1 x plus c2 you apply the boundary condition
y0 is 0 will give me c1 into 0 plus 2 so that
will give me c2 is 0 okay So y x becomes c1
x now you apply the other boundary condition
y at 1 minus y – at 1 is 0 So this will
give me c1 minus c1 this derivative that is
also c1 which is 0 it is satisfying So you
see that you take any arbitrary value of c1
so that means nonzero value of c1 it is actually
satisfying the 2nd boundary condition which
is satisfied okay
So that means so this is something like see
what you have is c1 times 1 minus 1 equal
to 0 So you have c1 times 0 is 0 That means
c1 is arbitrary Once you have c1 arbitrary
the general solution is this one So you have
nonzero solution so this implies lambda equal
to 0 is an eigenvalue And we call this lambda
0 corresponding to v equal to 0 So v0 of x
corresponding eigen function i am denoting
as v0 which is we can take c1 as 1 so you
have this is x is corresponding eigen function
Now look at the case lambda negative that
means lambda equal to minus mu square with
mu positive In this case what is your y double
dash minus mu square y equal to 0 this is
how the equation becomes general solution
of this equation is c1 e power mu x plus c2
e power minus qx You apply the boundary conditions
now we will give me c1 plus c2 equal to 0
that will give me c1 equal to minus c2 This
implies solution becomes general solution
becomes c1 times e power mu x minus c2 i am
replacing with minus c1 so you have e power
minus mu x Okay
So now you apply this is actually equal to
c1 2 c1 and this is sin hyperbolic mu x mu
is positive okay So now you apply the other
boundary condition y at 1 minus y – at 1
equal to 0 will give me for this we are 2
c1 sin hyperbolic mu minus 2 c1 mu cos hyperbolic
mu equal to 0 So this will give me 2 c1 2
cannot be 0 so this has to be c1 times sin
hyperbolic mu minus mu cos hyperbolic mu has
to be 0 So this i just have to check whether
for some positive mu values this quantity
is 0 So again we do the same thing so to check
this sin hyperbolic mu minus mu cos hyperbolic
mu equal to 0 that is we can do if this is
tan hyperbolic mu equal to mu if cos hyperbolic
mu is not equal to 0 because you are dividing
with it
But this will never be 0 this will never be
0 for any mu positive value Actually you can
do it okay it is not just is the cos since
this is nonzero you can always divide Do you
look at this tan hyperbolic plot and this
y equal to mu plot This is y equal to mu and
you plot tan hyperbolic mu they never touch
anywhere okay You can just do it they do not
touch any other place So you can plot it and
see tan hyperbolic mu do not have any solution
nonzero solution Okay positive solution strictly
positive mu you do not have any solution
Actually touching and above it goes simply
goes above only touching at 0 this is the
only route but that is mu equal to 0 So implies
for a nonzero mu no nonzero no nonzero mu
satisfying tan hyperbolic mu equal to mu That
means that implies for mu positive this quantity
sin hyperbolic mu minus mu cos hyperbolic
mu is nonzero That means by looking at this
c has to be 0 c1 has to be 0
That implies the general solution becomes
earlier after applying the 1st boundary condition
over general solution is this now that you
found c1 is 0 that means this is completely
0 between 0 to 1 That means lambda negative
is not an eigenvalue Any lambda negative is
not an eigenvalue So what you found is finally
you have only you have this is one eigen function
corresponding to 0 and the other eigenvalues
and eigen functions which form mu i’s sin
mu i’s mu i is satisfying than mu equal
to mu So now you know what are your eigenvalues
and eigenvectors as usual you can write the
fourier transforms and fourier series in this
case
So any piecewise piecewise continuous function
fx which is defined between 0 to 1 is written
as fx which is equal to what you have it is
running from i is running from 0 to infinity
1 to infinity is what you have for lambda
positive and i lambda equal to 0 corresponding
to i equal to 0 So you have c i and what is
your functions vi so those are sin mu ix and
x you have 2 such things So you have you can
rewrite like c0 into x that is one corresponding
to lambda equal to 0 and these things you
can sum it up with n is from 1 to infinity
and you have cn sin mu nx Okay
So you have these are your eigen functions
sin mu ix so i will just change the indx as
n Okay Where ci so how do you get your c0
c0 you should get it from by this is your
eigen function this is your eigen function
so by multiplying this eigen function x and
take the dot product with f you can get your
c0 x x Okay And this if you take your dot
product with sin mu nx later mu nx you write
mx you can get cm got up with sin mu mx sin
mu mx okay So what are these so you can think
of c0 is integral 0 to 1 i do not have so
the weight function is only 1 so it is usual
dot product fx into x dx between 0 to 1 into
the real part so does not matter
So you have to divide by integral x square
so you have to write 0 to 1 x square dx That
is x cube by 3 between 0 to 1 so which is
1 by 3 So you have 1 by 3 so you have total
3 So you finally get c0 as this one between
this And cm has integral 0 to 1 fx sin mu
x dx diverted by this because you do not know
what is exactly your mu m we cannot may not
be able you can actually find sin square mu
mx dx This you can calculate and put it this
is how you find this So these are your fourier
transform generalised fourier transforms and
this is your fourier series You can think
of fx as the signal you can split it into
discrete frequencies at these frequencies
you can have these solutions okay
These are your discrete frequencies so we
have 1 to 0 to infinity So divide by frequency
you split the signal since we have this fourier
transform you can get back your signal by
combining all of these discrete frequencies
okay in these functions So this is how you
can get eigenvalues and eigen functions So
here you got implicitly so you would not find
explicitly these eigenvalues and eigen functions
right So this is the example i will tell you
story So this is our sturm louisville theory
so what you will learn from this sturm louisville
theory is it is just the property of second-order
linear differential equation okay ordinary
differential equation
So using this you can actually develop what
is the fourier transform and fourier series
of a of a function defined on a finite interval
So if you think of this as a real line which
is a periodic function okay And that finite
interval is whatever is defined repeated everywhere
as periodically So such a thing you can represent
as the fourier series in terms of eigenvalues
and eigen functions Okay So another use of
this sturm louisville theory is you try to
extract these sturm louisville problem when
you solve partial differential equations in
a simpler domains that we will do in the future
videos okay
So when you are solving these partial differential
equations in a simpler domains such as rectangular
or circular or elliptical domains you may
have to convert you may have to extract the
main idea is to to how will you solve this
partial differential equation idea is to extract
if you can extract this sturm louisville problem
out of the boundary value problem whatever
you have for the partial differential equation
then you get all the solutions eigenvalues
and eigen functions of the corresponding whatever
you get extracted sturm louisville problem
using them you combine them you make a general
solution of the pde and get back your unknowns
And take a linear superposition because it
is a linear equation you can only solve linear
partial differential equations on a simpler
domains using this sturm louisville problems
That we will see in the next videos
