Good morning to everyone. So, we will be looking
into, if you remember, whatever we have done
in the last class, we have just looked into
the theorems in the continuous domains, and
the development will be... Basically, we are
we our aim is in this portion of the syllabus
is in order to solve the partial differential
equations; and we will be looking into the
only linear and homogenous partial differential
equations.
In the last few classes, we have looked into
- how we can classify and categorize the partial
differential equations into parabolic, elliptical,
and hyperbolic form; not only that, we have
also looked into the several boundary conditions
those are associated in order to solve a partial
differential equations; and we categorized
different types of boundary conditions as
well.
Now, in this class, what we will be looking
into? We will be formulating the standard
eigenvalue problem in continuous domain; earlier,
we have looked into - how these problems can
be formulated in discrete domain, whenever
we talked about the matrices; and that eigenvalue,
eigenvector method was utilized in order to
solve a system of algebraic equations or system
of ordinary differential equation. We will
be developing the theory for the continuous
domain in order to solve the partial differential
equation.
So, let us look into a particular form of
equation. Consider the equation 
in the form, d square y dx square plus lambda
y is equal to 0. Let us consider a second
order ordinary differential equation of this
particular form, subject to the boundary conditions,
at x is equal to 0, y is equal to 0; at x
is equal to 1, y is equal to 0. So, therefore,
we have selected a homogeneous equation with
homogeneous boundary conditions.
So, now clearly, in this equation since, the
equation and the governing equation and the
boundary conditions are clearly satisfied
by the solution y is equal to 0. So, therefore,
y is equal to 0, is a solution to this problem;
but this solution is not a solution that we
are looking for; this solution is known as
a trivial solution. So, y is equal to 0 is
of course, a solution, but it is a trivial
solution and we are not looking for that;
what we are looking for, is that, we are looking
for non-trivial solution.
So, we are basically looking for a solution,
non-zero solution, for different values of
lambda, so what are the values of lambda,
so that, these will be giving you a non-trivial
solution. So, if that exists, if there exists
a lambda that is a scalar, for which 
a non-trivial 
solution of y is obtained, then we call lambda
as an eigenvalue 
of the system; corresponding y is or corresponding
y or the solution is known as the eigen function.
So, one has to be very clear that one can
formulate an eigenvalue problem, if and only
if, the boundary conditions are homogeneous.
We can formulate an eigenvalue problem, if
and only if, for a completely homogenous system.
What do you mean by the homogeneous system?
By homogeneous system, we mean that the differential
equation is homogeneous; the boundary conditions
are also homogeneous.
If the differential equation and the boundary
conditions, both are homogeneous, then we
call that system as homogeneous system. Now,
for a completely homogeneous system, there
exists an eigenvalue problem; so, eigenvalue
problem means, first we have to check whether
the boundary conditions are homogeneous, as
well as the governing equations are homogeneous;
then only, we call that problem an eigenvalue
problem. And, if we if the governing equation
and the boundary conditions satisfy these
two basic criteria or property, then we should
go ahead for the solution for the eigenvalues
and corresponding eigen functions.
Now, for this particular problem, let us solve
this equation and see what are the eigenvalues
we are getting and also, the corresponding
eigen functions. So, we assume lambda as the
constant, lambda is real; if lambda is real,
then there are three possible values lambda
can assume: number 1 - lambda is equal to
0, so that is possibility one; Second part
is lambda is negative; let us say, it is minus
alpha square, so, this simply indicates lambda
is negative; and third is lambda positive,
this simply indicates lambda is positive.
So, let us consider these three cases separately
and see what you get. First case will be lambda
is equal to 0; if that is the case, let us
see - what is the form of our differential
equation. Now, the differential equation boils
down to d square y dx square is equal to 0;
and because lambda itself is 0, so we just
integrate it, and let us look into the solution
of this, we integrate this out; if we integrate
this equation, the first integration results
to dy dx is equal to some constant C 1 and
second integration leads to y as a function
of x is equal to C 1x plus C 2.
Now, these two constants - C 1and C 2 - should
be evaluated from the two boundary conditions,
whatever we have; let us write down the two
boundary conditions: at x is equal to 0, y
is equal to 0; that, if we put this boundary
condition here, let us see what we get, 0
is equal to C 1times 0 plus C 2. So, therefore,
C 2 is equal to 0; if C 2 is equal to 0, the
form of the equation has become, now, y is
equal C 1x. Now, let us put the other boundary
condition that at x is equal to 1, y is equal
to 0; so if we utilize this boundary condition,
let us see what we get.
If we put x is equal to 1 into y is equal
at y is equal to 0; if we look into the solution,
solution is, y is equal to C 1x and at x is
equal to 1, y is equal to 0 gives me an equation
C 1 times 1. So that simply implies that C
1is equal to 0. So, what we get? If you look
into solution, you will be getting y C 1x
plus C 2, both C 1and C 2 turns out to be
0; therefore y is equal to 0 is the solution
that we are getting. But this solution is
a trivial solution, therefore, one should
not so this trivial...
We are not looking for a trivial solution,
we are looking for a non-trivial solution;
so, therefore, lambda cannot be equal to 0.
So, lambda is equal to 0 is ruled out, so
we cannot use that anymore; so what is the
let us look into the second option. Second
option is lambda is negative, so, lambda is
equal to minus alpha square; so, lambda is
negative. Now, let us see, what is the form
of our differential equation under this condition
of lambda? Differential equation becomes d
square y dx square minus alpha square y is
equal to 0.
So, if you know the solution, form of this
solution is e to the power mx. So, the form
of the solution is in the form y is equal
to e to the power mx is the form of solution;
so if we do that, we just put y is equal to
e to the power mx and let us see what you
get. You will be getting a polynomial of order
2 of m and that equation is known as the characteristic
equation.
So, if we put it that way, so dy dx, if y
is e to the power mx, then dy dx is nothing
but m e to the power mx. We differentiate
it once more, so, it becomes d square y dx
square is equal to m square e to the power
mx. Now, we put all these value in the differential
equation, differential equation is d square
y dx square minus alpha square y is equal
to 0. If we put this here, m square e to the
power mx minus alpha square; y is e to the
power mx, so, it will be e to power mx is
equal to 0.
Now, m being not equal to 0, so, e to the
power mx is always positive; so, m square
minus alpha square is the solution; so, m
will be having a solution - plus minus alpha.
So, therefore, the form of the solution will
be in this form; so this is known as the characteristic
equation. So, form of the equation will be
solution will be C 1 e to the power alpha
x plus C 2 e to the power minus alpha x. So,
m will be having two roots, m 1 and m 2; so
form of the solution will be C 1 e to the
power m 1 x plus C 2 e to the power m 2 x;
in m 1, we put plus alpha; in m 2, we put
minus alpha.
So, now, with this solution, we put the boundary
conditions. So, let us write form of the solution
as C 1 e to the power alpha x plus C 2 e to
the power minus alpha x.
Next, we write down the boundary conditions:
at x is equal to 0, y is equal to 0; at x
is equal to 1, y is equal to 0, the homogeneous
boundary conditions. Now, if you utilize the
first boundary conditions, let us see what
we get. We will be getting 0 is equal to C
1, e to the power 0 is 1, plus C 2; so, C
1 is equal to minus C 2. So, the solution
is, we put, C 1 is minus C 2, so, minus C
2 e to the power alpha x plus C 2 e to the
power minus alpha x; so minus C 2 we can take
it common, so e to the power alpha x minus
e to the power minus alpha x. Now, we put
the other boundary condition and see, what
is the fate of y of the constant C 2? So,
one constant has to be this constant has to
be evaluated and one boundary condition is
left.
So, we will be having 0, minus C 2 e to the
power 1 minus e to the power minus e to the
power alpha minus e to the power minus alpha.
Now, if you remember the variation of e to
the power alpha as a function of alpha, it
will be... alpha is equal to 0; we have ruled
out alpha is equal to 0, because if alpha
is equal to 0, we are going to get a trivial
solution. So, even for alpha is equal to 0,
e to the power alpha is 1, from 1 onwards,
for any positive value of alpha, it will be
ever increasing function; it will be always
positive.
Similarly, if you look into the... So, this
is the variation of e to the power alpha;
I am plotting the variation of e to the power
alpha and e to the power minus alpha as a
function of alpha. So, e to the power alpha
is always positive; the minimum value is 1
and the maximum value will be always positive,
and the higher values will be always positive.
If you look into the value of e to the power
minus alpha, for alpha is equal to 0, e to
the power minus alpha is 1, and it decreases
exponentially for e to the power minus alpha,
minus infinity it will be 0, but, so, it varies
from 1 to 0; so, therefore e to the power
minus alpha is always positive.
So, what I mean by this analysis is that,
e to the power alpha is ever positive, e to
the power minus alpha is always positive;
so, their combination is always a positive
quantity. So, in order to satisfy this equation
only option is that C 2 is equal to 0; now,
if C 2 is equal to zero, from this, we will
be getting C 1 is equal to 0; so, what is
the fate of my solution? The fate of my solution
is 0 multiplied by e to the power alpha x
plus, C 2 is 0, multiplied by e to the power
minus alpha x, so the full solution becomes
0.
So, again we are landing up with a trivial
solution, so, this is nothing but a trivial
solution; and we are not looking for a trivial
solution, we are looking for a non-trivial
solution. So, therefore, alpha so therefore
lambda cannot be a negative quantity, it is
also not possible. So, lambda is minus alpha
is also not possible. So, we have if we remember,
we have three choices for real value of lambda;
lambda is 0, that is ruled out, because you
are getting a trivial solution; then lambda
is negative, that is also ruled out, because
you are getting a trivial solution. So, only
option that we are leaving with is, only lambda
is positive; and let us see what we get.
Case number 3 - lambda is positive and it
is equal to plus alpha square. So, d square
y dx square plus alpha square y is equal to
0. So, let us again look into the solution,
the form of the solution is again y is equal
to e to the power mx; so, we just put the
double differentiation of this equation here,
so this becomes m square e to the power mx
plus alpha square e to the power mx is equal
to 0.
So, what we get is, m square plus alpha square
e to the power mx is equal to 0; and e to
the power mx for non-zero value of m and x,
it will be ever positive; so, m square is
equal to minus alpha square. So, what we get
is that, what is the solution? The solution
is of course, minus root over plus minus root
over e to the power minus alpha square; so
it will be root over minus 1, it will be nothing
but i, imaginary quantity; so, it will be
plus minus i alpha.
So that is the solution of this characteristic
equation and the two roots are i alpha and
minus i alpha. So, let us see, how the solution
of the differential equation is now, takes
the form. So, y is equal to a function of
x; this is equal to C 1 e to the power m 1
x plus C 2 e to the power m 2 x, so, the form
of the solution is C 1 e to the power i alpha
x plus C 2 e to the power minus i alpha x;
and if you know, you can write the then the
Euler's equation; thus, e to the power i alpha
is nothing but cosine alpha plus i sin alpha.
Now, if you write, e to the minus i alpha
is nothing but cosine alpha minus i sin alpha.
If you open up these two quantities and write
it down in this equation, finally, you will
be getting the solution in the form of periodic
functions; this cosine and sin functions are
periodic functions and it will be combination
of the sin function and the cosine function.
So, final form of the solution becomes C 1
sin alpha x plus C 2 cosine alpha x; these
imaginary quantities i s, etcetera, will be
consumed in the constants C 1 and C 2. Now,
let us put the boundary conditions: at x is
equal to 0, y is equal to 0. So, let us see
at x is equal to 0, y equal to 0, so, put
y is equal 0; then, C 1 sin 0 plus C 2 cosine
0; our sin 0 is always 0, cos 0 is 1; so,
0 is equal to... So, 0 multiply C 1 is 0,
so, this be 1; so, C 2 is equal to 0; so this
boundary condition gives me the solution as
C 2 is equal to 0. So, what is the solution
of y x? It is nothing but C 1 sin alpha x.
Now, let us put the other boundary condition
and see what we get as the solution of this
equation. Other boundary condition is: at
x is equal to 1, y is equal to 0, so, if you
put y is equal to 0 C 1 sin alpha. Now, there
are two options of getting the solution; either
C 1 is equal to 0 or sin alpha is equal to
0; if we get, if we use C 1 is equal to 0,
then again, we are going to land up with a
trivial solution.
So, if C 1 is equal to 0, we get a trivial
solution; so, that is ruled out. So, we are
not looking for a trivial solution; we are
looking for a non-trivial solution. So, therefore,
C 1 cannot be is equal to 0; so what is the
option left? The option is left as sin alpha
is equal to 0. If you look into the solution,
generic solution of sin alpha is that the
alpha in this solution of this equation is,
alpha n is equal to n pi, where your n s,
are varying from 1, 2, 3 up to infinity.
Now, n is equal to 0 is ruled out; because,
if you put n is equal to 0, then alpha n is
alpha will be equal to 0; if alpha is equal
to 0, then again, we are going to get a trivial
solution. So, for getting a non-trivial solution,
the solution is alpha n is equal to n pi,
where the index n runs from 1 to infinity.
Now, for each alpha, for each value of n,
the corresponding values of alpha are known
as the eigen values. So, these are nth eigenvalue
of the system and what is the corresponding
eigen function? We denote the... So, what
is, if you look into the solution, what is
the solution?
The solution was C 1 sin alpha x; so, corresponding
to nth eigen value, we will be getting the
nth eigen function; we denote it as a y subscript
n. So, this becomes C 1 sin alpha n; so, this
becomes alpha n x; and, alpha n become n pi,
so it becomes n pi x; so this is the nth eigen
function. So, the corresponding eigen function
is sin n pi x and corresponding eigen value
is n pi.
So, next, what we will do? We just change
the boundary condition from Dirichlet boundary
conditions. So, what do we have? We have the
boundary conditions - homogeneous boundary
conditions: at x is equal to 0, y is equal
to 0, and at x is equal to 1, y is equal to
0; these are the Dirichlet boundary conditions.
Now, let us see, what is the form of eigen
values and eigen functions in the case of
a Neumann boundary condition.
So, again, we look into the problem in a different
way by changing the boundary condition. The
form of the equation remains same, d square
y dx square is equal to plus lambda y is equal
to 0. Now, what do we do? We change the boundary
condition; one boundary conditions from Dirichlet
to Neumann, so, subject to, at x is equal
to 0, we have a boundary condition dy dx is
equal to 0; and, at x is equal to 1, we have
y is equal to 0.
So, what is the difference between this problem
and earlier problem? The governing equation
remains the same, only the Dirichlet boundary
condition at x is equal to 0 is replaced by
a Neumann boundary condition dy dx is equal
to 0; and at the other end, at x is equal
to 1, the boundary condition remains the same;
it is a Dirichlet boundary condition. So that
is the difference between this problem and
other problem, so it is a... But you please
remember, notice that the ordinary differential
equation and the boundary conditions, both
are homogenous; only the nature of boundary
condition has been changed at the surface
or at the boundary x is equal to 0.
So, we note this down, both differential equation
and boundary conditions, they are homogeneous;
only the boundary condition has been changed.
So, again there will be... We will solve this
three cases completely, so, there will be
three options for real alpha; for real values
of alpha, there will be three cases; case
1: alpha is equal to 0; case 2: alpha is negative.
So, let us say, that was lambda right, so,
this one was lambda, so we put lambda is equal
to 0; lambda is minus alpha square, so, lambda
is negative; and the third option is lambda
is equal to plus alpha square, so it is positive.
Now, we examine all these three cases one
after another. So, case 1 will be lambda equal
to 0, so, what we will be getting is that,
d square y dx square is equal to 0. Now, we
have already seen earlier, the solution of
this equation is y is equal to C 1 x plus
C 2. Next, we put the boundary conditions,
utilize these two boundary conditions, and
evaluate the constants C 1 and C 2 and see
what we get.
Now, if you really do that, what we will be
getting is that solution is, C 1 x plus C
2 at x is equal to 0, d y dx is equal to 0.
So, if we evaluate dy dx of this equation,
what we get is dy dx is nothing but C 1; so,
at x is equal to 0, dy dx is equal to 0; use
that boundary condition, so, what you get
is, C 1 is equal to 0. So, therefore, the
solution is C 1 is equal to 0, so you will
be getting y is equal to C 2; so, therefore,
the solution is a constant solution. Now,
use the other boundary condition, that is,
at x is equal to 1, y is equal to 0; so, there
is no variation of x with respect to y. Since
y is constant, in order to satisfy this boundary
condition, then you have y is equal to 0;
C 2 is equal to C 2.
So, if you remember that whatever I have told
in the last class, that differential equation
must be valid throughout the whole boundary,
as well as throughout the whole control volume,
as well as it is valid only at the boundaries;
but the boundary conditions are valid only
at the boundaries. So, let us note down this
point and use it; so, differential equation
is valid at any point inside the control volume,
as well as on the boundary; on the other hand,
the boundary conditions are valid only at
boundaries, they need not be valid within
the control volume.
So, therefore, my differential equation must
be satisfying every point inside the control
volume, as well as, it must be satisfying
the boundary condition. Now, if we... So,
we have seen that solution of the differential
equation is a constant and that constant is
equal to C 2; but the differential equation
must be equal to 0, at the boundary x is equal
to 1. So, these two can go hand in hand, if
and only if, we put y is equal to 0 is equal
to C 2. So, therefore, C 2 becomes 0; and
if you look into the solution, C 1 is 0, C
2 is 0, so you will be getting y is equal
to 0 is a solution; and if you remember, this
is a trivial solution.
So, what is our conclusion? Our conclusion
is that alpha where lambda cannot be negative;
lambda cannot be minus alpha square, so that
is ruled out. So, what we have done till now?
We have ruled out, that lambda cannot be equal
to 0; lambda cannot be is equal to minus alpha
square, because both of these leads to only
trivial solution; and we are looking into
the non-trivial solutions. So, the third option
is now left; the lambda is positive and it
is equal to plus alpha square.
So, let us again look into the solution d
square y dx square plus lambda y is equal
to 0, so, put lambda is equal to alpha square;
so, you will be getting d square y dx square
plus alpha square y is equal to 0. So, again
e to the power mx is the form of the solution
and we have already solved this equation earlier;
I am not going to solve this equation once
again here. So, if you look into the characteristic
equation, the characteristic equation will
be in the form of m square plus alpha square
is equal to 0, so you will be having two roots;
m 1, 2 is nothing but plus minus i alpha.
So, therefore, the solution will be composed
of sine functions and cosine functions, periodic
functions. So, y is equal y as a function
of x becomes C 1 sin alpha x plus C 2 cosine
alpha x; that is the form of the solution.
And, let us put the boundary conditions and
evaluate C 1 and C 2; the first boundary is:
at x is equal to 0, dy dx is equal to 0. So,
if we put this boundary condition and let
us see what is our solution. y as a function
of x is nothing but C 1 sin alpha x plus C
2 cosine alpha x. Now, we evaluate dy dx,
we evaluate dy dx; dy dx becomes C 1 alpha
cos alpha x minus C 2 alpha sin alpha x, remembering
that differentiation of sin alpha x is nothing
but alpha cosine alpha x, and differentiation
of cosine alpha x nothing but minus alpha
sin alpha x.
Now, putting x is equal to 0, dy dx is equal
to 0, so what you will be getting is, 0 is
equal to C 1 alpha cos 0 is one minus C 2
alpha 0. So, you will be getting C 1 times
alpha is equal to 0; alpha is not equal to
0, because if alpha is equal to 0, you will
be landing with a trivial solution again;
and we are not looking for a trivial solution,
that means, C 1 is equal to 0; if C 1 is equal
to 0, let us see what we get as a solution.
So, the solution we will be getting is, y
as a function of x; since C 1 equal to 0,
so you will be getting C 2 cosine alpha x.
So, that is the form of the solution we are
getting, and this solution is a non-trivial
solution.
Now, we have to check; we have to evaluate
the eigenvalues and eigen functions. So, we
have one more condition that is left behind,
that is, the boundary condition at x is equal
to 1. So, at x is equal to 1, we have a Dirichlet
boundary condition there, so we put y is equal
to y x as 0, and this becomes C 2 cosine alpha;
we put x is equal to alpha. Again, in this
equation, C 2 cannot be equal to 0, why? If
C 2 is equal to 0, then again, we are going
back to a trivial solution and we are not
looking for a trivial solution. So, what is
the option left? The option left is cosine
alpha is equal to 0 for a non-trivial solution.
If cosine alpha is equal to 0, then let us
look into what is the general solution for
this equation. We have already studied the
general solution of sin alpha equal to 0,
cosine alpha is equal to 0, in our 10, plus
2 standard. So, the general form of this equation
is alpha n is 2 n minus 1 pi by 2, where the
index n runs from 1, 2 up to infinity. So,
this is the general form of the solution;
now, each of these values of alpha for which
cosine alpha is equal to 0 is the eigenvalues
for this particular problem. So, alpha n is
equal to 2 n minus 1 pi by 2 is the nth eigenvalue
of this problem; and the corresponding solution
is the n th eigen function; so, y n as a function
of x is nothing but... This is nth eigen function
and this becomes cosine alpha n x. So, therefore,
this becomes cosine 2 n minus 1 pi by 2 into
x.
So, alpha n is equal to 2 n minus 1 pi by
2 is the nth eigenvalue and y n cosine 2 n
minus 1 pi by 2 times x is the nth eigen function;
so, this is the eigenvalue; this is the eigen
function. So, if you look into the solution
for this particular problem by changing the
boundary condition from Dirichlet to Neumann,
we land up with a different eigenvalue and
eigen function combination. Next, what do
we do? We will change the boundary condition
to the third kind of boundary, which are most,
which are quite common in chemical engineering
applications, that is a Robin mixed boundary
condition. So, we change the boundary condition,
we keep one boundary condition - Dirichlet;
and like this particular case, we change the
boundary condition to the other boundary condition
to Robin mixed boundary condition and see,
what are the eigenvalues and eigen functions,
we are getting in this case.
So, let us consider the same equation d square
y dx square plus lambda y is equal to 0, but
we change the boundary conditions in this
case, subject to this is: at x is equal to
0, y is equal to 0. We keep the Dirichlet
boundary condition here, as it is; we change
the other boundary condition at x is equal
to 1, we use a Neumann boundary we use a Robin
mixed boundary condition, so, that will be
dy dx plus beta y is equal to 0. Please check
that, again we have a homogeneous system of
equation, the differential equation and boundary
conditions are all homogeneous.
What we simply did was, we keep this Dirichlet
boundary condition as it is; we change the
boundary condition to the other boundary condition,
that is, an x equal to 1, we make it a Robin
mixed boundary condition, where beta is a
constant.
So, therefore, what we get is that we look
into the again we this lambda can be for real
values of lambda. It can be three things;
lambda can be 0, lambda can be negative and
third option is lambda is positive. Let us
examine all the three cases in detail and
see what kind of eigen values and what kind
of eigen functions, we are getting in this
particular case.
So, for lambda, case 1 is lambda is equal
to 0 and again, the form of differential equation
is d square y dx square is equal to 0 and
we know the solution to this problem and the
solution becomes y is equal to C 1 x plus
C 2. Now, let us put the boundary conditions;
the first boundary condition is: at x is equal
to 0, y is equal to 0; if we put that you
will be getting y is equal to C 1 into 0;
so it will be 0. So, y is equal to 0 is equal
to C 2; so C 2 will be equal to 0. So, what
is the solution? The solution is, y is equal
to C 1 x.
Now, if we use the other boundary condition
that is at x is equal to 0, d y d x plus beta
y is equal to 0. So, we put: at x is equal
to 0, dy dx what? dy dx is C 1, and we put
this here; so, C 1 at x is equal to 1 at x
is equal to 1 dy dx plus beta y is equal to
0, that is the other boundary condition. So,
boundary conditions as defined on two boundaries,
at x is equal to 0 and x is equal to 1.
So, this becomes dy dx is C 1 plus beta times
y; so, beta times y evaluated at x is equal
to 1; so, this will be beta C 1 times 1 should
be is equal to 0; so, C 1 into 1 plus beta
is equal to 0. Now, beta is a constant; it
is a positive constant, so therefore, 1 plus
beta is always positive; so, 1 plus beta always
positive, it cannot be equal to 0; so, beta
can be positive or it can be negative; so,
therefore, 1 plus beta is not equal to 0.
So, in order to satisfy this equation, C 1
must be equal to 0; so if C 1 is equal to
0, so what is the form of our solution? This
is the solution we had; so, y is equal to
C 1 is 0, so y is equal to 0. So, y is equal
to 0, we are getting as a final solution for
this case as well; so, this is giving a trivial
solution. And, we are not looking for a trivial
solution, so therefore, lambda cannot be equal
to 0 in order to get a non-trivial solution.
So, in order to get a non-trivial solution
lambda is not equal to 0, so what is the next
option? The next option is, lambda is negative
and let us see what we get.
So, case 2: lambda is minus alpha square and
it is negative; so, our differential equation
now becomes d square y dx square minus alpha
square y is equal to 0. If you remember, the
form of the solution to this equation is e
to the power mx, and m square is equal to
minus alpha square is the characteristic equation.
So, the solution will be composed of e to
the power y is equal to e to the power alpha
x plus C 2 e to the power minus alpha x.
Now, let us put the first boundary condition;
the first boundary condition is: at x is equal
to 0, you have y is equal to 0. So, therefore,
this will be 0, C 1 plus C 2; so C 1 e to
the power 0 is one and e to the power minus
0 is also 1, so it will be C 1 plus C 2. So,
therefore, we will be getting C 2 is equal
to minus C 1, so the form of the solution
becomes y as a function of x, so, C 1 C 2
is equal to minus C 1; so, take C 1 common,
e to the power alpha x minus e to the power
minus alpha x.
So, next what we do? We utilize the other
boundary condition; the other boundary condition
is at x is equal to 1, we have dy dx plus
beta y is equal to 0; so let us evaluate.
So, if you look into the solution, the solution
is y as a function of x is C 1 e to the power
alpha x minus e to the power minus alpha x;
so we evaluate dy dx and see what we get.
dy dx is C 1 alpha e to the power alpha x
minus minus plus alpha e to the power minus
alpha x, so, we can take alpha common; so
C 1alpha e to the power alpha x plus e to
the power minus alpha x.
Now, we put the boundary condition, at x is
equal to 1; so dy dx will be evaluated, at
x is equal to 1 e to the power alpha x plus
e to the power minus alpha x; the whole term
is evaluated at x is equal to 1 plus beta
times y, y is C 1 e to the power alpha x minus
e to the power minus alpha x, whole term evaluated
at x is equal to 1 should be is equal to 0.
So, we evaluate that; so this becomes C 1alpha
e to the power alpha plus e to the power minus
alpha plus beta times C 1 e to the power alpha
minus e to the power minus alpha is equal
to 0. So, we can take C 1common; so this becomes
alpha e to the power alpha plus e to the power
minus alpha plus beta e to the power alpha
minus e to the power minus alpha is equal
to 0. Now, as we have argued earlier, that
we have looked into the variation of e to
the power alpha and e to the power minus alpha
with respect to alpha; this is the variation
of e to the power alpha, and that is the variation
of e to the power minus alpha. So, therefore,
e to the power alpha is always positive having
a minimum value 1; e to the power minus alpha
is always positive with a maximum value as
1.
So, their combination is positive again, their
combination is positive; beta is not equal
to 0; the whole thing in the third bracket
is a positive quantity that means so in order
to satisfy this equation, we have C 1 is equal
to 0. And let us see, if C 1is equal to 0
and if we look into the solution, the solution
was y is equal to C 1 e to the power alpha
x minus e to the power minus alpha x. For
C 1is equal 0 of course, you will be getting
y is equal to 0, which is nothing but a trivial
solution.
So, therefore, we are going to get a trivial
solution in this case as well. So, our conclusion
is, our lambda cannot be a negative quantity,
because it gives a trivial solution and we
are looking for a non-trivial solution. So,
I stop here, at the in this class at this
point; I will take this point on in the next
class and we will be completing this problem
in the next class and then we will take a
stop of what are the forms of eigenvalues
and eigen functions if we change the boundary
conditions slightly.
Thank you very much.
