Differential Equations - Fourier Series: Eigenvalues and Eigenfunctions - i

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 $\lambda$ for which we could find nonzero solutions, i.e. $\overrightarrow{x}\ne \overrightarrow{0}$, to,

$$A\overrightarrow{x}=\lambda \overrightarrow{x}$$

then we called $\lambda$ an eigenvalue of $A$ and $\overrightarrow{x}$ was its corresponding eigenvector.

It’s important to recall here that in order for $\lambda$ 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,

$$\begin{array}{}{y}^{\prime \prime }+\lambda y=0y\left(0\right)=0y\left(2\pi \right)=0& \text{(1)}\end{array}$$

In Example 7 we had $\lambda =4$ and we found nontrivial (i.e. nonzero) solutions to the BVP. In Example 8 we used $\lambda =3$ and the only solution was the trivial solution (i.e. $y\left(t\right)=0$). 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 $\lambda$that will give nontrivial solutions to this BVP and values of $\lambda$ that will only admit the trivial solution.

So, for those values of $\lambda$ that give nontrivial solutions we’ll call $\lambda$ 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 $\text{(1)}$ $\lambda =4$ is an eigenvalue (with eigenfunctions $y\left(x\right)=c_2\sin \left(2x\right)$) and that $\lambda =3$ is not an eigenvalue.

Eventually we’ll try to determine if there are any other eigenvalues for $\text{(1)}$, 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

$$y^{\prime \prime }+4y=0$$

with two different nonhomogeneous boundary conditions in the form,

$$y\left(0\right)=ay\left(2\pi \right)=b$$

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 equation and its characteristic polynomial has two real, distinct roots and that they are in the form

$$r_1=\alpha r_2=-\alpha$$

Then we know that the solution is,

$$y\left(x\right)=c_1{e}^{r_{1}x}+c_2{e}^{r_{2}x}=c_1{e}^{\alpha x}+c_2{e}^{-\alpha x}$$

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,

$$\begin{array}{cc}y\left(x\right)& =c_1{e}^{\alpha x}+c_2{e}^{-\alpha x}\\ & =\frac{c_1}{2}{e}^{\alpha x}+\frac{c_1}{2}{e}^{\alpha x}+\frac{c_2}{2}{e}^{-\alpha x}+\frac{c_2}{2}{e}^{-\alpha x}\end{array}$$

Now we’ll add/subtract the following terms (note we’re “mixing” the $c_i$ and $\pm \alpha$ up in the new terms) to get,

$$y\left(x\right)=\frac{c_1}{2}{e}^{\alpha x}+\frac{c_1}{2}{e}^{\alpha x}+\frac{c_2}{2}{e}^{-\alpha x}+\frac{c_2}{2}{e}^{-\alpha x}+\left(\frac{c_1}{2}{e}^{-\alpha x}-\frac{c_1}{2}{e}^{-\alpha x}\right)+\left(\frac{c_2}{2}{e}^{\alpha x}-\frac{c_2}{2}{e}^{\alpha x}\right)$$

Next, rearrange terms around a little,

$$y\left(x\right)=\frac{1}{2}\left(c_1{e}^{\alpha x}+c_1{e}^{-\alpha x}+c_2{e}^{\alpha x}+c_2{e}^{-\alpha x}\right)+\frac{1}{2}\left(c_1{e}^{\alpha x}-c_1{e}^{-\alpha x}-c_2{e}^{\alpha x}+c_2{e}^{-\alpha x}\right)$$

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,

$$\begin{array}{cc}y\left(x\right)& =\left(c_1+c_2\right)\frac{e^{\alpha x}+e^{-\alpha x}}{2}+\left(c_1-c_2\right)\frac{e^{\alpha x}-e^{-\alpha x}}{2}\\ & ={\overline{c}}_1\frac{e^{\alpha x}+e^{-\alpha x}}{2}+{\overline{c}}_2\frac{e^{\alpha x}-e^{-\alpha x}}{2}\end{array}$$

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,

$$\cosh \left(x\right)=\frac{e^x+e^{-x}}{2}\sinh \left(x\right)=\frac{e^x-e^{-x}}{2}$$

So, another way to write the solution to a second order differential equation whose characteristic polynomial has two real, distinct roots in the form $r_1=\alpha ,r_2=-\alpha$ is,

$$y\left(x\right)=c_1\cosh \left(\alpha x\right)+c_2\sinh \left(\alpha x\right)$$

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,

$$\frac{d}{dx}\left(\cosh \left(x\right)\right)=\sinh \left(x\right)\frac{d}{dx}\left(\sinh \left(x\right)\right)=\cosh \left(x\right)$$

Next let’s take a quick look at the graphs of these functions.

Note that $\cosh \left(0\right)=1$ and $\sinh \left(0\right)=0$. Because we’ll often be working with boundary conditions at $x=0$ these will be useful evaluations.

Next, and possibly more importantly, let’s notice that $\cosh \left(x\right)>0$ for all $x$ and so the hyperbolic cosine will never be zero. Likewise, we can see that $\sinh \left(x\right)=0$ only if $x=0$. 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.$$y^{\prime \prime }+\lambda y=0y\left(0\right)=0y\left(2\pi \right)=0$$

We started off this section looking at this BVP and we already know one eigenvalue ($\lambda =4$) and we know one value of $\lambda$that is not an eigenvalue ($\lambda =3$). As we go through the work here we need to remember that we will get an eigenvalue for a particular value of $\lambda$ if we get non-trivial solutions of the BVP for that particular value of $\lambda$.
In order to know that we’ve found all the eigenvalues we can’t just start randomly trying values of $\lambda$ to see if we get non-trivial 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 : $\lambda >0$, $\lambda =0$, and $\lambda <0$. 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 non-trivial solutions or not. So, let’s get started on the cases.
$\underset{–}{\lambda >0}$
In this case the characteristic polynomial we get from the differential equation is,
$$r^2+\lambda =0\Rightarrow r_{1,2}=\pm \sqrt{-\lambda }$$In this case since we know that $\lambda >0$ these roots are complex and we can write them instead as,
$$r_{1,2}=\pm \sqrt{\lambda }i$$The general solution to the differential equation is then,
$$y\left(x\right)=c_1\cos \left(\sqrt{\lambda }x\right)+c_2\sin \left(\sqrt{\lambda }x\right)$$Applying the first boundary condition gives us,
$$0=y\left(0\right)=c_1$$So, taking this into account and applying the second boundary condition we get,
$$0=y\left(2\pi \right)=c_2\sin \left(2\pi \sqrt{\lambda }\right)$$This means that we have to have one of the following,
$$c_2=0\text{or}\sin \left(2\pi \sqrt{\lambda }\right)=0$$However, recall that we want non-trivial solutions and if we have the first possibility we will get the trivial solution for all values of $\lambda >0$. Therefore, let’s assume that $c_2\ne 0$. This means that we have,
$$\sin \left(2\pi \sqrt{\lambda }\right)=0\Rightarrow 2\pi \sqrt{\lambda }=n\pi n=1,2,3,\dots$$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 $\lambda >0$ we know that $2\pi \sqrt{\lambda }>0$and so $n$ can only be a positive integer for this case.
Now all we have to do is solve this for $\lambda$ and we’ll have all the positive eigenvalues for this BVP.
The positive eigenvalues are then,
$${\lambda }_n={\left(\frac{n}{2}\right)}^2=\frac{n^2}{4}n=1,2,3,\dots$$and the eigenfunctions that correspond to these eigenvalues are,
$$y_n\left(x\right)=\sin \left(\frac{nx}{2}\right)n=1,2,3,\dots$$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 $c_2$ 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.
$\underset{–}{\lambda =0}$
In this case the BVP becomes,
$$y^{\prime \prime }=0y\left(0\right)=0y\left(2\pi \right)=0$$and integrating the differential equation a couple of times gives us the general solution,
$$y\left(x\right)=c_1+c_{2}x$$Applying the first boundary condition gives,
$$0=y\left(0\right)=c_1$$Applying the second boundary condition as well as the results of the first boundary condition gives,
$$0=y\left(2\pi \right)=2c_2\pi$$Here, unlike the first case, we don’t have a choice on how to make this zero. This will only be zero if $c_2=0$.
Therefore, for this BVP (and that’s important), if we have $\lambda =0$ the only solution is the trivial solution and so $\lambda =0$cannot be an eigenvalue for this BVP.
Now let’s look at the final case.
$\underset{–}{\lambda <0}$
In this case the characteristic equation and its roots are the same as in the first case. So, we know that,
$$r_{1,2}=\pm \sqrt{-\lambda }$$However, because we are assuming $\lambda <0$ 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,
$$y\left(x\right)=c_1\cosh \left(\sqrt{-\lambda }x\right)+c_2\sinh \left(\sqrt{-\lambda }x\right)$$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,
$$0=y\left(0\right)=c_1\cosh \left(0\right)+c_2\sinh \left(0\right)=c_1\left(1\right)+c_2\left(0\right)=c_1\Rightarrow c_1=0$$Applying the second boundary condition gives,
$$0=y\left(2\pi \right)=c_2\sinh \left(2\pi \sqrt{-\lambda }\right)$$Because we are assuming $\lambda <0$ we know that $2\pi \sqrt{-\lambda }\ne 0$ and so we also know that $\sinh \left(2\pi \sqrt{-\lambda }\right)\ne 0$. Therefore, much like the second case, we must have $c_2=0$.
So, for this BVP (again that’s important), if we have $\lambda <0$ 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.
$${\lambda }_n=\frac{n^2}{4}{y}_n\left(x\right)=\sin \left(\frac{nx}{2}\right)n=1,2,3,\dots$$

Let’s take a look at another example with slightly different boundary conditions.