Eigenvalues and Eigenfunctions
As we did in the previous section we need to again note that
we are only going to give a brief look at the topic of eigenvalues and
eigenfunctions for boundary value problems.
There are quite a few ideas that we’ll not be looking at here. The intent of this section is simply to give
you an idea of the subject and to do enough work to allow us to solve some
basic partial differential equations in the next chapter.
Now, before we start talking about the actual subject of
this section let’s recall a topic from Linear Algebra that we briefly discussed previously in
these notes. For a given square matrix, A, if we could find values of for which we could find nonzero solutions, i.e. ,
to,
then we called an eigenvalue of A and was its corresponding eigenvector.
It’s important to recall here that in order for to be an eigenvalue then we had to be able to
find nonzero solutions to the equation.
So, just what does this have to do with boundary value
problems? Well go back to the previous
section and take a look at Example
7 and Example 8. In those two examples we solved homogeneous
(and that’s important!) BVP’s in the form,
In Example 7 we had and we found nontrivial (i.e. nonzero) solutions to the BVP.
In Example 8 we used and the only solution was the trivial solution
(i.e. ).
So, this homogeneous BVP (recall this also means the boundary conditions
are zero) seems to exhibit similar behavior to the behavior in the matrix
equation above. There are values of that will give nontrivial solutions to this
BVP and values of that will only admit the trivial solution.
So, for those values of that give nontrivial solutions we’ll call an eigenvalue
for the BVP and the nontrivial solutions will be called eigenfunctions for the BVP corresponding to the given eigenvalue.
We now know that for the homogeneous BVP given in (1)
is an eigenvalue (with eigenfunctions ) and that is not an eigenvalue.
Eventually we’ll try to determine if there are any other
eigenvalues for (1),
however before we do that let’s comment briefly on why it is so important for
the BVP to be homogeneous in this discussion.
In Example 2
and Example 3 of the
previous section we solved the homogeneous differential equation
with two different nonhomogeneous boundary conditions in the
form,
In these two examples we saw that by simply changing the
value of a and/or b we were able to get either nontrivial
solutions or to force no solution at all.
In the discussion of eigenvalues/eigenfunctions we need solutions to
exist and the only way to assure this behavior is to require that the boundary
conditions also be homogeneous. In other
words, we need for the BVP to be homogeneous.
There is one final topic that we need to discuss before we
move into the topic of eigenvalues and eigenfunctions and this is more of a
notational issue that will help us with some of the work that we’ll need to do.
Let’s suppose that we have a second order differential
equations and its characteristic polynomial has two real, distinct roots and
that they are in the form
Then we know that the solution is,
While there is nothing wrong with this solution let’s do a
little rewriting of this. We’ll start by
splitting up the terms as follows,
Now we’ll add/subtract the following terms (note we’re
“mixing” the and up in the new terms) to get,
Next, rearrange terms around a little,
Finally, the quantities in parenthesis factor and we’ll move
the location of the fraction as well.
Doing this, as well as renaming the new constants we get,
All this work probably seems very mysterious and
unnecessary. However there really was a
reason for it. In fact you may have
already seen the reason, at least in part.
The two “new” functions that we have in our solution are in fact two of
the hyperbolic functions. In particular,
So, another way to write the solution to a second order
differential equation whose characteristic polynomial has two real, distinct
roots in the form is,
Having the solution in this form for some (actually most) of
the problems we’ll be looking will make our life a lot easier. The hyperbolic
functions have some very nice properties that we can (and will) take advantage
of.
First, since we’ll be needing them later on, the derivatives
are,
Next let’s take a quick look at the graphs of these
functions.
Note that and . Because we’ll often be working with boundary
conditions at these will be useful evaluations.
Next, and possibly more importantly, let’s notice that for all x
and so the hyperbolic cosine will never be zero. Likewise, we can see that only if . We will be using both of these facts in some
of our work so we shouldn’t forget them.
Okay, now that we’ve got all that out of the way let’s work
an example to see how we go about finding eigenvalues/eigenfunctions for a BVP.
Example 1 Find
all the eigenvalues and eigenfunctions for the following BVP.
Solution
We started off this section looking at this BVP and we
already know one eigenvalue ( ) and we know one value of that is not an eigenvalue ( ).
As we go through the work here we need to remember that we will get an
eigenvalue for a particular value of if we get nontrivial solutions of the BVP
for that particular value of .
In order to know that we’ve found all the eigenvalues we
can’t just start randomly trying values of to see if we get nontrivial solutions or
not. Luckily there is a way to do this
that’s not too bad and will give us all the eigenvalues/eigenfunctions. We are going to have to do some cases
however. The three cases that we will
need to look at are : ,
,
and . Each of these cases gives a specific form
of the solution to the BVP to which we can then apply the boundary conditions to
see if we’ll get nontrivial solutions or not. So, let’s get started on the cases.
In this case the characteristic
polynomial we get from the differential equation is,
In this case
since we know that these roots are complex and we can write
them instead as,
The general
solution to the differential equation is then,
Applying the
first boundary condition gives us,
So, taking this
into account and applying the second boundary condition we get,
This means that
we have to have one of the following,
However, recall
that we want nontrivial solutions and if we have the first possibility we
will get the trivial solution for all values of . Therefore let’s assume that . This means that we have,
In other words,
taking advantage of the fact that we know where sine is zero we can arrive at
the second equation. Also note that
because we are assuming that we know that and so n can only be a positive integer for this case.
Now all we have
to do is solve this for and we’ll have all the positive eigenvalues
for this BVP.
The positive
eigenvalues are then,
and the eigenfunctions
that correspond to these eigenvalues are,
Note that we
subscripted an n on the eigenvalues
and eigenfunctions to denote the fact that there is one for each of the given
values of n. Also note that we dropped the on the eigenfunctions. For eigenfunctions we are only interested
in the function itself and not the constant in front of it and so we
generally drop that.
Let’s now move
into the second case.
In this case
the BVP becomes,
and integrating
the differential equation a couple of times gives us the general solution,
Applying the
first boundary condition gives,
Applying the
second boundary condition as well as the results of the first boundary
condition gives,
Here, unlike
the first case, we don’t have a choice on how to make this zero. This will only be zero if .
Therefore, for
this BVP (and that’s important), if we have the only solution is the trivial solution
and so cannot be an eigenvalue for this BVP.
Now let’s look
at the final case.
In this case
the characteristic equation and its roots are the same as in the first
case. So, we know that,
However,
because we are assuming here these are now two real distinct
roots and so using our work above for these kinds of real, distinct roots we
know that the general solution will be,
Note that we
could have used the exponential form of the solution here, but our work will
be significantly easier if we use the hyperbolic form of the solution here.
Now, applying
the first boundary condition gives,
Applying the
second boundary condition gives,
Because we are
assuming we know that and so we also know that . Therefore, much like the second case, we
must have .
So, for this
BVP (again that’s important), if we have we only get the trivial solution and so
there are no negative eigenvalues.
In summary then we will have the following
eigenvalues/eigenfunctions for this BVP.

Let’s take a look at another example with slightly different
boundary conditions.
Example 2 Find
all the eigenvalues and eigenfunctions for the following BVP.
Solution
Here we are going to work with derivative boundary
conditions. The work is pretty much
identical to the previous example however so we won’t put in quite as much
detail here. We’ll need to go through
all three cases just as the previous example so let’s get started on that.
The general
solution to the differential equation is identical to the previous example
and so we have,
Applying the
first boundary condition gives us,
Recall that we
are assuming that here and so this will only be zero if . Now, the second boundary condition gives
us,
Recall that we
don’t want trivial solutions and that so we will only get nontrivial solution if
we require that,
Solving for and we see that we get exactly the same
positive eigenvalues for this BVP that we got in the previous example.
The
eigenfunctions that correspond to these eigenvalues however are,
So, for this
BVP we get cosines for eigenfunctions corresponding to positive eigenvalues.
Now the second
case.
The general
solution is,
Applying the
first boundary condition gives,
Using this the
general solution is then,
and note that
this will trivially satisfy the second boundary condition,
Therefore,
unlike the first example, is an eigenvalue for this BVP and the
eigenfunctions corresponding to this eigenvalue is,
Again, note
that we dropped the arbitrary constant for the eigenfunctions.
Finally let’s
take care of the third case.
The general
solution here is,
Applying the
first boundary condition gives,
Applying the
second boundary condition gives,
As with the
previous example we again know that and so . Therefore we must have .
So, for this
BVP we again have no negative eigenvalues.
In summary then we will have the following
eigenvalues/eigenfunctions for this BVP.
Notice as well
that we can actually combine these if we allow the list of n’s for the first one to start at zero
instead of one. This will often not
happen, but when it does we’ll take advantage of it. So the “official” list of
eigenvalues/eigenfunctions for this BVP is,

So, in the previous two examples we saw that we generally
need to consider different cases for as different values will often lead to
different general solutions. Do not get
too locked into the cases we did here.
We will mostly be solving this particular differential equation and so
it will be tempting to assume that these are always the cases that we’ll be
looking at, but there are BVP’s that will require other/different cases.
Also, as we saw in the two examples sometimes one or more of
the cases will not yield any eigenvalues.
This will often happen, but again we shouldn’t read anything into the
fact that we didn’t have negative eigenvalues for either of these two
BVP’s. There are BVP’s that will have
negative eigenvalues.
Let’s take a look at another example with a very different
set of boundary conditions. These are
not the traditional boundary conditions that we’ve been looking at to this
point, but we’ll see in the next chapter how these can arise from certain
physical problems.
Example 3 Find
all the eigenvalues and eigenfunctions for the following BVP.
Solution
So, in this example we aren’t actually going to specify
the solution or its derivative at the boundaries. Instead we’ll simply specify that the
solution must be the same at the two boundaries and the derivative of the solution
must also be the same at the two boundaries.
Also, this type of boundary condition will typically be on an interval
of the form [L,L] instead of [0,L] as we’ve been working on to this point.
As mentioned above these kind of boundary conditions arise
very naturally in certain physical problems and we’ll see that in the next
chapter.
As with the previous two examples we still have the
standard three cases to look at.
The general
solution for this case is,
Applying the
first boundary condition and using the fact that cosine is an even function (i.e. ) and that sine is an odd function (i.e. ).
gives us,
This time,
unlike the previous two examples this doesn’t really tell us anything. We could have but it is also completely possible, at this
point in the problem anyway, for us to have as well.
So, let’s go
ahead and apply the second boundary condition and see if we get anything out
of that.
So, we get
something very similar to what we got after applying the first boundary
condition. Since we are assuming that this tells us that either or .
Note however
that if then we will have to have and we’ll get the trivial solution. We therefore need to require that and so just as we’ve done for the previous
two examples we can now get the eigenvalues,
Recalling that and we can see that we do need to start the
list of possible n’s at one instead
of zero.
So, we now know
the eigenvalues for this case, but what about the eigenfunctions. The solution for a given eigenvalue is,
and we’ve got
no reason to believe that either of the two constants are zero or nonzero
for that matter. In cases like these
we get two sets of eigenfunctions, one corresponding to each constant. The two sets of eigenfunctions for this
case are,
Now the second
case.
The general
solution is,
Applying the
first boundary condition gives,
Using this the
general solution is then,
and note that
this will trivially satisfy the second boundary condition just as we saw in
the second example above. Therefore we
again have as an eigenvalue for this BVP and the
eigenfunctions corresponding to this eigenvalue is,
Finally let’s
take care of the third case.
The general
solution here is,
Applying the
first boundary condition and using the fact that hyperbolic cosine is even
and hyperbolic sine is odd gives,
Now, in this
case we are assuming that and so we know that which in turn tells us that . We therefore must have .
Let’s now apply
the second boundary condition to get,
By our
assumption on we again have no choice here but to have .
Therefore, in
this case the only solution is the trivial solution and so, for this BVP we
again have no negative eigenvalues.
In summary then we will have the following
eigenvalues/eigenfunctions for this BVP.
Note that we’ve
acknowledged that for we had two sets of eigenfunctions by listing
them each separately. Also, we can
again combine the last two into one set of eigenvalues and eigenfunctions. Doing so gives the following set of
eigenvalues and eigenfunctions.

Once again we’ve got an example with no negative
eigenvalues. We can’t stress enough that
this is more a function of the differential equation we’re working with than
anything and there will be examples in which we may get negative eigenvalues.
Now, to this point we’ve only worked with one differential
equation so let’s work an example with a different differential equation just
to make sure that we don’t get too locked into this one differential
equation.
Before working this example let’s note that we will still be
working the vast majority of our examples with the one differential equation
we’ve been using to this point. We’re
working with this other differential equation just to make sure that we don’t
get too locked into using one single differential equation.
Example 4 Find
all the eigenvalues and eigenfunctions for the following BVP.
Solution
This is an Euler differential
equation and so we know that we’ll
need to find the roots of the following quadratic.
The roots to
this quadratic are,
Now, we are going to again have some cases to work with
here, however they won’t be the same as the previous examples. The solution will depend on whether or not
the roots are real distinct, double or complex and these cases will depend
upon the sign/value of . So, let’s go through the cases.
In this case
the roots will be complex and we’ll need to write them as follows in order to
write down the solution.
By writing the
roots in this fashion we know that and so is now a real number, which we need in order
to write the following solution,
Applying the
first boundary condition gives us,
The second
boundary condition gives us,
In order to
avoid the trivial solution for this case we’ll require,
This is much
more complicated of a condition than we’ve seen to this point, but other than
that we do the same thing. So, solving
for gives us the following set of eigenvalues
for this case.
Note that we
need to start the list of n’s off
at one and not zero to make sure that we have as we’re assuming for this case.
The
eigenfunctions that correspond to these eigenvalues are,
Now the second
case.
In this case we
get a double root of and so the solution is,
Applying the
first boundary condition gives,
The second
boundary condition gives,
We therefore
have only the trivial solution for this case and so is not an eigenvalue.
Let’s now take
care of the third (and final) case.
This case will
have two real distinct roots and the solution is,
Applying the
first boundary condition gives,
Using this our
solution becomes,
Applying the
second boundary condition gives,
Now, because we
know that for this case the exponents on the two terms
in the parenthesis are not the same and so the term in the parenthesis is not
the zero. This means that we can only
have,
and so in this
case we only have the trivial solution and there are no eigenvalues for which
.
The only
eigenvalues for this BVP then come from the first case.

So, we’ve now worked an example using a differential
equation other than the “standard” one we’ve been using to this point. As we saw in the work however, the basic
process was pretty much the same. We
determined that there were a number of cases (three here, but it won’t always
be three) that gave different solutions.
We examined each case to determine if nontrivial solutions were
possible and if so found the eigenvalues and eigenfunctions corresponding to
that case.
We need to work one last example in this section before we
leave this section for some new topics.
The four examples that we’ve worked to this point were all fairly
simple (with simple being relative of course…), however we don’t want to leave
without acknowledging that many eigenvalue/eigenfunctions problems are so easy.
In many examples it is not even possible to get a complete
list of all possible eigenvalues for a BVP.
Often the equations that we need to solve to get the eigenvalues are
difficult if not impossible to solve exactly.
So, let’s take a look at one example like this to see what kinds of things
can be done to at least get an idea of what the eigenvalues look like in these
kinds of cases.
Example 5 Find
all the eigenvalues and eigenfunctions for the following BVP.
Solution
The boundary conditions for this BVP are fairly different
from those that we’ve worked with to this point. However, the basic process is the
same. So let’s start off with the
first case.
The general
solution to the differential equation is identical to the first few examples
and so we have,
Applying the
first boundary condition gives us,
The second
boundary condition gives us,
So, if we let we’ll get the trivial solution and so in
order to satisfy this boundary condition we’ll need to require instead that,
Now, this
equation has solutions but we’ll need to use some numerical techniques in
order to get them. In order to see
what’s going on here let’s graph and on the same graph. Here is that graph and note that the
horizontal axis really is values of as that will make things a little easier to
see and relate to values that we’re familiar with.
So, eigenvalues
for this case will occur where the two curves intersect. We’ve shown the first five on the graph and
again what is showing on the graph is really the square root of the actual
eigenvalue as we’ve noted.
The interesting
thing to note here is that the farther out on the graph the closer the
eigenvalues come to the asymptotes of tangent and so we’ll take advantage of
that and say that for large enough n
we can approximate the eigenvalues with the (very well known) locations of
the asymptotes of tangent.
How large the
value of n is before we start using
the approximation will depend on how much accuracy we want, but since we know
the location of the asymptotes and as n
increases the accuracy of the approximation will increase so it will be easy
enough to check for a given accuracy.
For the
purposes of this example we found the first five numerically and then we’ll
use the approximation of the remaining eigenvalues. Here are those values/approximations.
The number in
parenthesis after the first five is the approximate value of the
asymptote. As we can see they are a
little off, but by the time we get to the error in the approximation is
0.9862%. So less than 1% error by the
time we get to and it will only get better for larger value
of n.
The
eigenfunctions for this case are,
where the
values of are given above.
So, now that
all that work is out of the way let’s take a look at the second case.
The general
solution is,
Applying the
first boundary condition gives,
Using this the
general solution is then,
Applying the
second boundary condition to this gives,
Therefore for
this case we get only the trivial solution and so is not an eigenvalue. Note however that had the second boundary
condition been then would have been an eigenvalue (with
eigenfunctions ) and so again we need to be careful
about reading too much into our work here.
Finally let’s
take care of the third case.
The general
solution here is,
Applying the
first boundary condition gives,
Using this the
general solution becomes,
Applying the
second boundary condition to this gives,
Now, by
assumption we know that and so . This in turn tells us that and we know that for all x. Therefore,
and so we must
have and once again in this third case we get the
trivial solution and so this BVP will
have no negative eigenvalues.
In summary the only eigenvalues for this BVP come from
assuming that and they are given above.

So, we’ve worked several eigenvalue/eigenfunctions examples
in this section. Before leaving this
section we do need to note once again that there are a vast variety of
different problems that we can work here and we’ve really only shown a bare
handful of examples and so please do not walk away from this section believing
that we’ve shown you everything.
The whole purpose of this section is to prepare us for the
types of problems that we’ll be seeing in the next chapter. Also, in the next chapter we will again be
restricting ourselves down to some pretty basic and simple problems in order to
illustrate one of the more common methods for solving partial differential
equations.