In the last two lectures, we started discussions
on the modal analysis of continuous systems.
Now, this was the performance of modal analysis
was found to be essentially solving an Eigen
value problem. Now, today we are going to
look at some properties of this Eigen value
problem that comes up while we perform modal
analysis of continuous systems.
So, let us start by re-visiting the modal
analysis problem. So, in our last lecture,
we had discussed the problem of a bar with
varying cross section. The equation of motion
of the system was 
given by this and the relevant boundary conditions
for this problem were given by U at zero is
equal to zero for all time and on the right
boundary we have natural boundary condition
and in the order to do the modal analysis
we were searching for solutions of the special
form or structure.
So, the field variable is expressed as a product
of an amplitude function which is the function
of x, and harmonically varying time function.
Now, we discuss the properties of this solution
and we found that the solution is actually
separable in space and time. So, when we write
the actual solution in real form… it appears
in this structure. So, it was separable in
space and time. The other observation is all
points, therefore, vibrate at the same frequency
omega, the same circular frequency omega.
Thirdly, all points of the system pass through
the equilibrium point at the same time instant.
The time instant when this temporal function
is zero, the whole solution is zero, which
means the bar is in its equilibrium state;
so, all points will pass through the equilibrium
point at the same time.
Then we observed that phase difference between
any two points and the bar is either zero
or pi. And finally, we observe the existence
of modes that means points at which U, the
amplitude function capital U of x is zero.
So, the points, so, the properties of the
modal solution are known to us. So, once we
substitute the solution of this structure
in to the equation of motion, we obtain 
the differential equation in terms of this
amplitude function and the corresponding 
boundary conditions. This forms the Eigen
value problem for the system, so, the differential
equation along with the boundary conditions.
Now, we will represent this in a slightly
abstract form 
in this manner… where, our equation of motion
was, can be written like this…
So, if you write a general equation of motion
of a string or a bar in this form, then the
differential equation of the Eigen value problem
may be represented in this manner, where lambda
is omega square. So, this actually is plus.
So, this is the differential equation and
the corresponding differential equation for
the Eigen value problem is given in this form,
where lambda is omega square and this K is
the differential operator.
So, for example, in the case of the tapered
bar, mu(x) is rho times the area 
and the differential operator K, which is
also known as the stiffness operator, is the
spatial derivative of this quantity, E A in
the derivative of the argument. So, this is
the structure of the differential equation
of our Eigen value problem. Now here, as I
mentioned, here this is known as the stiffness
operator, because this term comes from the
potential energy in the Lagrangian formulation,
while this term mu(x) is the kinetic energy
operator, because it comes from the kinetic
energy in the Lagrangian formulation.
So, suppose for two modes j and k we can write
this differential equation. So, for the jth
mode let us say, we can write the differential
equation of the Eigen value problem like this,
while for the kth mode, the differential equation
becomes this. Now, we are going to, the objective
of this analysis is to determine certain properties
of the Eigen value problem. So, let me multiply
this first equation with Wk and the second
equation with Wj, and subtract one from the
other, then after some rearrangement, I can
write. So, I am also integrating over the
domain of the system.
So, I multiplied the first equation with Wk
and the second equation with Wj subtracted
one from the other and integrated over the
domain of the problem; and this is what I
obtain upon rearrangement. Now, suppose that
this integral vanishes. So, let us consider
the situation when this property holds 
where this W and W tilde are functions that
satisfy 
the boundary conditions of the problem. If
this property holds, then this operator K
is known as self-adjoint. So, this property
is satisfied by the stiffness operator then
it is known as a self-adjoint operator.
Now, this self adjointness of an operator
is connected to symmetry. So, as you know
that the stiffness operator has a corresponding
matrix, for example, in vibrations of discrete
systems you come across stiffness matrix.
The self adjointness of the stiffness operator
is nothing but the symmetry of the stiffness
matrix, the corresponding stiffness matrix.
So, what are the consequences of this symmetry?
As we know that when matrices are symmetric,
the Eigen values are real and the Eigen functions
are also real, and the Eigen vectors are orthogonal.
So, in a similar manner, we have these properties
which can be shown very easily that the Eigen
values are real, Eigen values and Eigen functions
are real, whenever the stiffness operator
is self-adjoint. Secondly, the Eigen functions
are orthogonal 
with respect to an inner product that we will
find out in the course of this lecture.
So, we are, we will be now discussing this
orthogonality property. So, re-call that we
have this equation. So, if the operator K
is self-adjoint then this term vanishes. So,
this implies this integral must vanish, whenever
j is not equal to k. So, which means if I
take two distinct amplitude functions modes,
Eigen functions Wj and Wk, then this satisfy
this property that this integral must vanish;
and this, we define as 
the inner product of these two Eigen functions.
And in a compact form, we will write this,
the inner product of two Eigen functions can
be written in this form where alpha j 
is given by this integral.
Now, one may normalize this property by appropriately
scaling the Eigen functions, because as we
know that, any scaled form of this Eigen function
is also an Eigen function. So, we can scale
appropriately, to have ortho-normality of
the Eigen functions with respect to this inner
product that we have defined. So, here 
this Wj hat is Wj over square root of alpha
j. So, here we have orthogonality with respect
to the inertia operator. So, if you consider
that this mu of x represents the inertia operator,
then this orthogonality is with respect to
the inertia operator and correspondingly we
have we can write… So, for the jth mode,
this is the differential equation. So, if
I multiply this equation with… So, this
can be written also for the hat Eigen function,
the normalized Eigen function and if I multiply
this with Wk hat and integrate… So, this
shows that the Eigen functions are orthogonal
also with respect to the stiffness operator
K. Now, what is, what are the implications
of this, so, what are what is the physical
implication of 
this orthogonality with respect to the inertia
and the stiffness operators. So, the implication
is that, there is no exchange of kinetic or
potential energy between the Eigen modes.
The physical implication is that there is
no exchange of kinetic or potential energy
between the Eigen modes. And this orthogonality
property is also very useful as we will see
in due course for solving initial value problems
or other problems related to continuous systems.
And this orthogonality we have already come
across when we discussed about vibrations
of modal analysis of strings.
Next, we discuss some examples 
and we will determine the orthogonality relations
for these examples. So, we once again go back
to this bar of varying cross-section and follow
the steps that we have done in detail. So,
our Eigen value problem… Now, let us check
that this stiffness operator, that we have
here, is really self-adjoint. So, what we
have to show… So, this we have to show.
So, you have to show that these two are equal
where Uk and Uj are two Eigen functions of
this Eigen value problem. So, we start integrating
by parts, let see from the left hand side.
So, we take this as the first function and
this as the second function. This is what
we obtain. And here I will integrate by parts
this term once again and here I will use the
boundary conditions. So, the boundary terms
here that I have, so, this term will be evaluated
at l and at 0. Now, at l U prime at l must
be 0. So, this term must vanish at l and U(0)
is 0. So, Uk at 0 must be 0 becomes these
are Eigen functions and satisfy the boundary
conditions of the Eigen value problems. So,
this term is actually zero. So, we are left
with only this term and this I will integrate
by parts once again.
Now, the same reasoning as we had here. This
term, this boundary terms must also varnish.
So, we are left with which is nothing but
the right hand side of this equation. So,
we have shown that 
this operator acting on… So, we have shown
this self adjointness of the stiffness operator
of that equal to bar. So, we can write this
orthogonality in terms of the inner product
as we were defining 
for the tapered bar.
Next, we are going to look at the hanging
string or hanging chain. So, for the hanging
chain the Eigen value problem 
write this. So, in this case, the stiffness
operator 
is 
given by this term and in a similar manner
you can check 
that this self adjointness property holds
for the stiffness operator of a hanging chain.
And once you, I mean, used this property,
you can derive the inner product of the Eigen
functions of the hanging chain with respect
to which the Eigen functions are orthogonal.
So, let me just write down this Eigen function
that we have already derived 
in a previous lecture. So… So, this is the
structure of the Eigen functions of a hanging
chain and they satisfy 
the orthogonality relation in this form with...
So, this is alpha j is the square of the Eigen
say that jth Eigen function and integrated
over 0 to l, and this turns out to be… However,
this J1 is the Basel function of order 1.
Now, let us consider the example of a uniform
bar 
which is coupled to a harmonic oscillator,
which we have discuss in our previous lecture.
So, the Eigen value problem 
for this system was written as… as obtain
in our previous lecture. So, once again for
mode j and mode k, we can… So, this is the
two differential equations and this is the
boundary conditions for the bar. So, for the
mode j and k, we can write…
Now, we will once again multiply 
this with the first equations for the bar
with Uk and the second with Uj subtract and
integrate over the domain of the bar, and
upon rearrangement, you can very easily obtain…
So, there are few standard steps. So, to obtain
from here to here, that we can easily perform
and come to this condition. Now, if you integrate
by parts, let say this first term. So, integrate
by parts this first term two times and used
the boundary conditions for the boundary terms
that you generate, then you can check that
the expression reduces … So, we are integrating
this term by parts two times. So at the end
of the integration by parts, this term will
be exactly same as this; so that two cancel
off, but we will generate two boundary terms
with single prime and that we have to replace
that, we have to use this boundary condition.
Once you use that you ultimately come to this
expression. Now when j is not equal to k,
and considering that omega j is not equal
to omega k, there are no repeated Eigen frequencies,
then this bracketed quantity must vanish;
and this if you check, this can be written
as…
Here I have replaced these quantities by Yj
and Yk which we have obtained this expression
before. So, this is our inner product. Remember
that in the case of a discrete, of a hybrid
system in which we have a continuous and discrete
system, we had this Eigen function vector
which we have discussed in the previous lecture.
So, here we would say, we will write the inner
product in this form, where the inner product
is now defined in this form. So, you see this
is not a trivial or a simple inner product
that we obtain for other systems. So here
we, so, this procedure you have to follow
in order to determine this structure of the
inner products, how this inner product is
calculated based on the Eigen functions. So,
let us summarize what we have studied today.
So, we had…
So, we have revisited this modal analysis
problem and the Eigen value problem. Then
we looked at the properties of the modal solution.
Then we discussed about self-adjoint 
operators and the consequences of the stiffness
operator being self-adjoint, these are real
Eigen values and real Eigen functions. Then
we have discussed about the orthogonality
property of Eigen functions 
and we have determined the inner product.
We have outlined steps to determine the inner
product with respect to which this orthogonality
property holds and we have looked at the implications
of the orthogonality property of the Eigen
functions. So, if the Eigen functions are
orthogonal that implies that there is no exchange
of energy, kinetic or potential, between the
Eigen modes or the Eigen functions. So, with
that we conclude this lecture.
Ketword: Eigen value problem, self-adjointness,
orthogonality of Eigen functions, inner product.
