We now know that
for the basic Fourier series theory
we need the following three eigenvalue problems.
In the last two lessons
we solve problems in the form of 4.1 and 4.2.
In this lesson, we'll determine the eigenvalues
and eigenfunctions of a problem in the form of 4.3.
Again, recall that a number lambda is an eigenvalue
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.
The following problem is the one
that leads to the general Fourier series.
We have x double prime plus lambda x equals zero
where x of negative pi equals x of pi
and x prime of negative pi equals x prime of pi.
We want to determine the eigenvalues and eigenfunctions.
Notice here we have not specified
the values or derivatives at endpoints,
but rather that they are the same
at the beginning and end of the interval,
as usually we need to consider lambda less than zero,
lambda equals zero and lambda greater than zero separately.
Let's begin with lambda less than zero.
We've done this two times before.
Recall if lambda is less than zero
if we multiply both sides by -1
we have negative lambda greater than zero.
And because the characteristic equation
has two distinct real roots
the general solution can be written 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.
We also need to find x prime
and therefore we differentiate with respect to t
which requires a chain rule shown here on the right.
And now I begin with the condition x of negative pi
equals x of pi.
We substitute negative pi for t in x
and then we substitute pi for t in x
and then set the two expressions equal to each other.
Analyzing this equation
and using the hyperbolic triggered entities shown below,
the hyperbolic cosine expressions
on both sides are equal, but the hyperbolic sine functions
on both sides are not equal
which indicates b must equal zero.
So if b is equal to zero, then we have x prime
is equal to a square root negative lambda
hyperbolic sine of square root negative lambda t.
And then we use a condition x prime of negative pi
equals x prime of pi.
So here we substitute negative pi for t in x prime
and then pi for t in x prime
and set the expressions equal to each other,
which is shown here on the right.
And this equation is only true when a equals zero.
So notice both a and b are equal to zero
and therefore when lambda is less than zero
the only solution is x equals zero
which indicates there are no negative eigenvalues.
Now consider lambda equals zero.
For lambda equals zero, we have the differential equation
x double prime equals zero
and therefore the general solution is
in the form of x equals At plus b, a linear function,
and x double prime is equal to zero.
Using the condition x of negative pi equals x of pi,
we have a times negative pi plus b
equals a times pi plus b.
Now for this to be true, notice a must equal zero.
If a is equal to zero
then we have the solution is x equals b.
And if x equals b, x prime is equal to zero
using the condition x prime of negative pi
equals x prime of pi
we do get zero equals zero, which is true
but this does not give us any new information
and therefore we have x equals b
for the corresponding eigenfunctions
with the eigenvalue, lambda equals zero.
So again, lambda equals zero is an eigenvalue
and x equals b is a corresponding eigenfunction
where b is any real number.
So if we let b equal 1
we can use the corresponding eigenfunction of x equals 1.
And now we consider lambda greater than zero.
For lambda greater than zero, the general solution
is in the form of x equals a cosine
of square root lambda t plus b sine square root lambda t.
And this is because the corresponding
characteristic equation is r squared plus lambda equals zero
giving us two complex roots
using the condition x of negative pi equals x of pi,
we get the equation shown here.
Recall that cosine negative theta equals cosine theta
and sine negative theta equals negative sine theta.
Therefore, begin write the equation
as a cosine square root of lambda pi minus b sine
square root of lambda pi
equals a cosine square root of lambda pi
plus b sine square root of lambda pi.
Notice cosine expressions are equal
on both sides of the equation
and the expressions involving the sine function
are opposites.
This indicates for the equation to be true
either b must be zero
or sine of square root of lambda pi must equal zero
and recall the sine function is zero
when the input is a multiple of pi.
This indicates in order
for the sine function to be zero
square root lambda must equal k
for any integers k greater than or equal to 1
or both a and b are zero.
But of course, we don't want both a and b to be zero.
So for a non-zero solution, we have lambda equals k squared
or k is greater than or equal to 1.
And now when you to find x prime
and use the condition
x prime of negative pi equals x prime of pi.
I've shown this below, we find that a equals zero
or sine square root of lambda pi equals zero
with the same result
that square root of lambda equals k
for any integers k greater than or equal to 1.
As a result once again, we get lambda equals k squared
or k is greater than or equal to 1.
So looking at the work below
here we have x and then x prime.
And the next line we have x prime of negative pi
equals x prime of pi.
Using our identities
we can write the equation as shown below
which again gives us square root of lambda equals k
for all integers k greater than or equal to 1
and therefore lambda equals k squared.
So using our general solution,
this indicates that x equals a cosine kt plus b sine kt.
Notice here we simply replaced square root of lambda with k.
So this is an eigenfunction for any values of a and b
which means we have two linearly independent eigenfunctions
where if we let a and b both equal 1
we can use the two linearly independent eigenfunctions.
x equals sine kt and x equals cosine kt.
Remember, a matrix can also have two eigenvectors
corresponding to a single eigenvalue
if the eigenvalue is repeated.
So in conclusion, when lambda is greater than zero
we have lambda equals k squared as our eigenvalues
where k is any integer greater than or equal to 1
and x equals cosine kt
and x equals sine kt are corresponding eigenfunctions.
Let's go ahead and summarize our findings.
So in summary, the eigenvalues
and corresponding eigenfunctions
are lambda sub k equals k squared
with eigenfunctions cosine kt and sine kt
for all integers k greater than or equal to 1.
And we have lambda subzero equals zero as an eigenvalue
with an eigenfunction of x subzero equals 1.
I hope you found this helpful.