In the last lesson we learned
for basic Fourier series
we need the following three eigenvalue problems.
In the last lesson, we solved an eigenvalue problem
in the form of 4.1.
In this lesson will solve an eigenvalue problem
in the form of 4.2.
Recalling 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.
In this lesson will compute the eigenvalues
And eigenfunctions of x double prime
plus lambda x equals zero with x prime of zero equals zero
and x prime of pi equals zero.
Again, we need to handle the cases lambda greater than zero,
lambda equals zero and lambda less than zero separately.
First suppose 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.
And this is because
the corresponding characteristic equation
is R squared plus lambda equals zero
where again lambda is greater than zero
and therefore the characteristic equation
has two complex solutions,
plus or minus i square root lambda.
So this is why the general solution is in the given form.
And because we have X prime of zero equals zero
and X prime of pi equals zero
we now need to find X prime,
which is derivative of x with respect to T shown here.
Notice this does require the chain rule.
And now we use a condition X prime of zero equals zero
which implies B equals zero.
I've shown this work in blue on the right.
Given X prime of zero equals zero,
we substitute zero for T into X prime
and set it equal to zero.
This gives us negative A square lambda sine zero
plus B square root lambda cosine zero equals zero.
We know sine zero is zero and cosine zero is one
which indicates the equation is only true
when B equals zero.
So if B is equal to zero, we now know X prime
is equal to negative A square root lambda
sine square root lambda T.
We also know X is equal to a cosine of square root lambda T.
And now using the condition X prime of pi equal zero
we substitute pi for T into X prime
and set it equal to zero.
Again using the equation X prime equals negative A
square root lambda sine square root lambda T,
we have negative A square root lambda
sine of square root lambda pi equals zero.
A can't be zero if lambda is an eigenvalue
and therefore the equation is true
when the sine square root lambda pi is zero,
and this only occurs when the input square root lambda pi
is a multiple of pi.
And therefore square root lambda must equal k,
for a positive integer k.
So square root lambda equals k,
and lambda equals k squared,
hence the positive eigenvalues are lambda equals k squared
for all integers k greater than or equal to one.
And the corresponding eigenfunctions
can be taken as X equals cosine kT.
Recall, we know that X equals A cosine square root lambda T.
And since square root lambda's equal to k
we have X equals A cosine kT.
If we let A equal one, we have X equals cosine kT.
And now we suppose that lambda equals zero.
In this case the equation becomes X double prime equals zero
and therefore the general solution
is a linear function X equals AT plus B.
Notice the second derivative is equal to zero
which does satisfy the differential equation.
From here, X prime is equal to A
using the condition X prime of zero equals zero.
Well, if X prime is equal to A
and X prime of zero equals zero,
we now know that A must equal zero.
And if A is equal to zero
then we have the equation X equals B.
Notice the condition X prime of pi equals zero
doesn't give us any new information
and still implies A equals zero.
So now we know B can be any real number.
So lambda equals zero is an eigenvalue
and X equals one is a corresponding eigenfunction.
So again, when lambda equals zero, lambda is an eigenvalue
and X equals B is a corresponding eigenfunction.
Where B is any real number, we will let B equal one
and we'll use the function X equals one
as a corresponding eigenfunction.
And now we consider lambda less than zero.
In this case, a 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.
This is a general solution
because the corresponding characteristic equation
when lambda is less than zero is R squared
minus lambda equals zero
or the roots R plus or minus square root lambda.
Recall in general, when the characteristic equation
has two distinct real roots,
for example, this equation here,
where the roots are plus or minus k
we can write the general solution
in the form of 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
in the form shown here
y equals D1 hyperbolic cosine kX
plus D2 hyperbolic sine kX.
So this explains why our general solution
involves the hyperbolic trig functions
and then we're using square root negative lambda because
if lambda is less than zero
and we multiply both sides by negative one,
this is equivalent to negative lambda
being greater than zero.
The next step we find X prime,
which requires the chain rule.
And then using the condition X prime of zero equals zero
we substitute zero for T into X prime
and set it equal to zero, which I've shown here in blue.
Hyperbolic sine zero is zero.
Hyperbolic cosine zero is one.
And therefore the equation is only true when B equals zero.
So if B is equal to zero, we now know X prime
is equal to A square root negative lambda
hyperbolic sine of square root negative lambda T.
Using the condition X prime of pi equals zero,
we substitute pi for T and set X prime equal to zero.
This gives us a square root negative lambda hyperbolic sine
of square root negative lambda pi equals zero.
Recall that hyperbolic sine is only zero
when the input is zero.
And because we're considering lambda less than zero
the input into the hyperbolic sine function is not zero,
which implies A equals zero.
So because both A and B are equal to zero
we don't have a non-zero solution
which indicates when lambda is less than zero
the only solution is X equals zero
and there are no negative eigenvalues.
Let's summarize our findings.
The eigenvalues and corresponding eigenfunctions are
lambda sub k equals k squared
with an eigenfunction, X sub k equals cosine kT
for all integers k greater than or equal to one.
And there is another eigenvalue lambda sub zero equals zero
with an eigenfunction X sub zero equals one.
I hope you found this helpful.