In this lesson we'll find the eigenvalues
and eigenfunctions of a given boundary value problem.
For basic Fourier series theory,
we'll need the following three eigenvalue problems.
For 4.1, we have X double prime plus lambda X equals zero,
where X of A and X of B equal zero.
For 4.2, we have X double prime plus lambda X equals zero
where X prime of A and X prime of B equal zero.
And for 4.3 we have X double prime plus lambda X equals zero
where X of A equals X of B,
and X prime of A equals X prime of B.
A number lambda is called an eigenvalue
of the boundary value problems 4.1 through 4.3
if and only if there exists a non-zero solution
to the problem given that specific lambda.
A non-zero solution is called a corresponding eigenfunction.
Note the similarity to eigenvalues and
eigenvectors of matrices.
The similarity is not coincidental.
If we think of the equations as differential operators
then we are doing the same exact thing.
Think of a function X of T as a vector
with infinitely many components, one for each T.
Let L equals the opposite of
the second order differential operator with respect to T
be the linear operator.
Then the eigenvalue-eigenfunction pair should be lambda
and a non-zero X such that LX equals lambda X.
In other words, we are looking for
non-zero functions X satisfying certain endpoint conditions
that solve the difference of
L and lambda times X equals zero.
Let's look at an example.
We will find the eigenvalues and eigenfunctions
of x double prime plus lambda X equals zero,
with X of zero equals zero, and X of pi equals zero.
We have to handle the cases lambda greater than zero,
lambda equals zero, and lambda less than zero, separately.
First suppose that lambda is greater than zero.
Then the general solution to
X double prime plus lambda X equals zero
is X equals A cosine of square root lambda T
plus B sine square root lambda T.
This is because the characteristic equation
is R squared plus lambda equals zero,
where the roots are plus or minus i square root lambda.
And now we consider the conditions.
First, we consider X of zero equals zero
which implies A equals zero.
I've shown this work on the right.
If X of zero equals zero,
then we have A cosine zero plus B sine zero equals zero.
Sine zero is zero, cosine zero is one,
which is why A must equal zero.
And now we know X of T equals B sine square root lambda T.
And now we consider the second condition of
X of pi equals zero
which gives us B sine square root Lambda pi equals zero.
If B is zero, then X is not a non-zero solution,
so to get a non-zero solution
we must have that sine of square root lambda pi equals zero.
Hence, square root lambda pi,
the input into the sine function
must be an integer multiple of pi because
the sine of an integer multiple of pi is equal to zero.
In other words, square root lambda must equal K
for a positive integer K.
So square root lambda equals K,
to solve for lambda, we square both sides
and we get lambda equals K squared.
Hence, the positive eigenvalues are K squared
for all integers K greater than or equal to one.
Corresponding eigenfunctions must then be taken as
X equals sine of KT, since square root lambda is equal to K.
Just like eigenvectors,
constant multiples of an eigenfunction
are also eigenfunctions, so we only need to pick one.
And now I move on to the second case,
where lambda equals zero.
Notice when lambda is equal to zero,
the differential equation simply becomes
X double prime equals zero.
So we're looking for the general form of a function
where the second derivative is zero,
which is a linear function in the form of
X equals AT plus B.
Notice if X equals AT plus B
X prime is equal to A, X double prime is equal to zero.
And now we consider our conditions.
First, we have X of zero equals zero.
If X of zero equals zero,
then A times zero plus B must equal zero
and therefore B is equal to zero.
Now we know X of T equals AT,
and if X of pi equals zero,
then A times pi must equal zero,
indicating A must also equal zero.
This means that lambda equals zero is not an eigenvalue.
Recall from our previous notes
the number lambda is called an eigenvalue if and only if
there exists a non-zero solution to
one of the boundary problems listed below.
And now we consider the third case
where lambda is less than zero.
Let's do this on the next slide.
For Lambda less than zero,
the general solution is in the form of X equals A
hyperbolic cosine of square root negative lambda T
plus B hyperbolic sine of square root negative lambda T.
If you're wondering where the general solution came from
recall from a previous example
if the characteristic equation has two real distinct roots
we can express a general solution as Y equals C1 e to the KX
plus C2 e to the negative KX.
But if we use the definitions of
hyperbolic cosine and hyperbolic sine,
we can also express a general solution as
Y equals D1 hyperbolic cosine KX plus D2 hyperbolic sine KX.
So going back to our problem,
you now may be wondering
why we have the square root of negative lambda
as part of the input for the hyperbolic trig functions.
We'll recall a lambda is less than zero
if we multiply both sides by negative one
this is equivalent to
negative lambda being greater than zero.
So this is why we're using negative lambda
under the square root.
And now we consider our conditions.
Again first, we have X of zero equals zero,
which I've shown below.
Recall that hyperbolic cosine zero was one
and hyperbolic sine zero was zero.
So performing this substitution,
hyperbolic sine zero is zero,
hyperbolic cosine zero is one,
which indicates A must equal zero,
and therefore we have
X of T equals B times
hyperbolic sine of square root negative lambda T.
And now we consider X of pi equals zero,
which indicates B equals zero.
If you're wondering why B is zero,
hyperbolic sine is zero only when the input is zero
which we can verify by looking at the graph of the
hyperbolic sign function, which is graphed here in red.
Notice it's only equal to zero when the input is zero.
We can also see this from the definition of hyperbolic sine.
We get zero equals hyperbolic sine xi equals
e to the xi minus e to the negative xi divided by two.
In order for this to be zero,
e to the xi must equal either negative xi,
which implies the exponents are equal,
meaning xi equals negative xi,
which is only true when xi is zero.
This indicates there are no negative eigenvalues.
In summary, the eigenvalues and corresponding eigenfunctions
are lambda sub K equals K squared,
with an eigenfunction X sub K equals sine KT
for all integers K greater than or equal to one.
I hope you found this helpful.