Differential Equations - Fourier Series: Eigenvalues and Eigenfunctions - ii

Example 2 Find all the eigenvalues and eigenfunctions for the following BVP.$$y^{\prime \prime }+\lambda y=0y^{\prime }\left(0\right)=0y^{\prime }\left(2\pi \right)=0$$

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.
$\underset{–}{\lambda >0}$
The general solution to the differential equation is identical to the previous example and so we have,
$$y\left(x\right)=c_1\cos \left(\sqrt{\lambda }x\right)+c_2\sin \left(\sqrt{\lambda }x\right)$$Applying the first boundary condition gives us,
$$0=y^{\prime }\left(0\right)=\sqrt{\lambda }{c}_2\Rightarrow c_2=0$$Recall that we are assuming that $\lambda >0$ here and so this will only be zero if $c_2=0$. Now, the second boundary condition gives us,
$$0=y^{\prime }\left(2\pi \right)=-\sqrt{\lambda }{c}_1\sin \left(2\pi \sqrt{\lambda }\right)$$Recall that we don’t want trivial solutions and that $\lambda >0$ so we will only get non-trivial solution if we require that,
$$\sin \left(2\pi \sqrt{\lambda }\right)=0\Rightarrow 2\pi \sqrt{\lambda }=n\pi n=1,2,3,\dots$$Solving for $\lambda$ and we see that we get exactly the same positive eigenvalues for this BVP that we got in the previous example.
$${\lambda }_n={\left(\frac{n}{2}\right)}^2=\frac{n^2}{4}n=1,2,3,\dots$$The eigenfunctions that correspond to these eigenvalues however are,
$$y_n\left(x\right)=\cos \left(\frac{nx}{2}\right)n=1,2,3,\dots$$So, for this BVP we get cosines for eigenfunctions corresponding to positive eigenvalues.
Now the second case.
$\underset{–}{\lambda =0}$
The general solution is,
$$y\left(x\right)=c_1+c_{2}x$$Applying the first boundary condition gives,
$$0=y^{\prime }\left(0\right)=c_2$$Using this the general solution is then,
$$y\left(x\right)=c_1$$and note that this will trivially satisfy the second boundary condition,
$$0=y^{\prime }\left(2\pi \right)=0$$Therefore, unlike the first example, $\lambda =0$ is an eigenvalue for this BVP and the eigenfunctions corresponding to this eigenvalue is,
$$y\left(x\right)=1$$Again, note that we dropped the arbitrary constant for the eigenfunctions.
Finally let’s take care of the third case.
$\underset{–}{\lambda <0}$
The general solution here is,
$$y\left(x\right)=c_1\cosh \left(\sqrt{-\lambda }x\right)+c_2\sinh \left(\sqrt{-\lambda }x\right)$$Applying the first boundary condition gives,
$$0=y^{\prime }\left(0\right)=\sqrt{-\lambda }{c}_1\sinh \left(0\right)+\sqrt{-\lambda }{c}_2\cosh \left(0\right)=\sqrt{-\lambda }{c}_2\Rightarrow c_2=0$$Applying the second boundary condition gives,
$$0=y^{\prime }\left(2\pi \right)=\sqrt{-\lambda }{c}_1\sinh \left(2\pi \sqrt{-\lambda }\right)$$As with the previous example we again know that $2\pi \sqrt{-\lambda }\ne 0$ and so $\sinh \left(2\pi \sqrt{-\lambda }\right)\ne 0$. Therefore, we must have $c_1=0$.
So, for this BVP we again have no negative eigenvalues.
In summary then we will have the following eigenvalues/eigenfunctions for this BVP.
$$\begin{array}{cccc}{\lambda }_n& =\frac{n^2}{4}& y_n\left(x\right)& =\cos \left(\frac{nx}{2}\right)n=1,2,3,\dots \\ {\lambda }_0& =0& y_0\left(x\right)& =1\end{array}$$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,
$${\lambda }_n=\frac{n^2}{4}{y}_n\left(x\right)=\cos \left(\frac{nx}{2}\right)n=0,1,2,3,\dots$$

So, in the previous two examples we saw that we generally need to consider different cases for $\lambda$ 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.$$y^{\prime \prime }+\lambda y=0y\left(-\pi \right)=y\left(\pi \right)y^{\prime }\left(-\pi \right)=y^{\prime }\left(\pi \right)$$

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.
$\underset{–}{\lambda >0}$
The general solution for this case is,
$$y\left(x\right)=c_1\cos \left(\sqrt{\lambda }x\right)+c_2\sin \left(\sqrt{\lambda }x\right)$$Applying the first boundary condition and using the fact that cosine is an even function (i.e.$\cos \left(-x\right)=\cos \left(x\right)$) and that sine is an odd function (i.e. $\sin \left(-x\right)=-\sin \left(x\right)$). gives us,
$$\begin{array}{cc}{c}_1\cos \left(-\pi \sqrt{\lambda }\right)+c_2\sin \left(-\pi \sqrt{\lambda }\right)& =c_1\cos \left(\pi \sqrt{\lambda }\right)+c_2\sin \left(\pi \sqrt{\lambda }\right)\\ c_1\cos \left(\pi \sqrt{\lambda }\right)-c_2\sin \left(\pi \sqrt{\lambda }\right)& =c_1\cos \left(\pi \sqrt{\lambda }\right)+c_2\sin \left(\pi \sqrt{\lambda }\right)\\ -c_2\sin \left(\pi \sqrt{\lambda }\right)& =c_2\sin \left(\pi \sqrt{\lambda }\right)\\ 0& =2c_2\sin \left(\pi \sqrt{\lambda }\right)\end{array}$$This time, unlike the previous two examples this doesn’t really tell us anything. We could have $\sin \left(\pi \sqrt{\lambda }\right)=0$ but it is also completely possible, at this point in the problem anyway, for us to have $c_2=0$ as well.
So, let’s go ahead and apply the second boundary condition and see if we get anything out of that.
$$\begin{array}{cc}-\sqrt{\lambda }{c}_1\sin \left(-\pi \sqrt{\lambda }\right)+\sqrt{\lambda }{c}_2\cos \left(-\pi \sqrt{\lambda }\right)& =-\sqrt{\lambda }{c}_1\sin \left(\pi \sqrt{\lambda }\right)+\sqrt{\lambda }{c}_2\cos \left(\pi \sqrt{\lambda }\right)\\ \sqrt{\lambda }{c}_1\sin \left(\pi \sqrt{\lambda }\right)+\sqrt{\lambda }{c}_2\cos \left(\pi \sqrt{\lambda }\right)& =-\sqrt{\lambda }{c}_1\sin \left(\pi \sqrt{\lambda }\right)+\sqrt{\lambda }{c}_2\cos \left(\pi \sqrt{\lambda }\right)\\ \sqrt{\lambda }{c}_1\sin \left(\pi \sqrt{\lambda }\right)& =-\sqrt{\lambda }{c}_1\sin \left(\pi \sqrt{\lambda }\right)\\ 2\sqrt{\lambda }{c}_1\sin \left(\pi \sqrt{\lambda }\right)& =0\end{array}$$So, we get something very similar to what we got after applying the first boundary condition. Since we are assuming that $\lambda >0$ this tells us that either $\sin \left(\pi \sqrt{\lambda }\right)=0$ or $c_1=0$.
Note however that if $\sin \left(\pi \sqrt{\lambda }\right)\ne 0$ then we will have to have $c_1=c_2=0$ and we’ll get the trivial solution. We therefore need to require that $\sin \left(\pi \sqrt{\lambda }\right)=0$ and so just as we’ve done for the previous two examples we can now get the eigenvalues,
$$\pi \sqrt{\lambda }=n\pi \Rightarrow \lambda =n^{2}n=1,2,3,\dots$$Recalling that $\lambda >0$ 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,
$$y\left(x\right)=c_1\cos \left(nx\right)+c_2\sin \left(nx\right)$$and we’ve got no reason to believe that either of the two constants are zero or non-zero 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,
$$y_n\left(x\right)=\cos \left(nx\right)y_n\left(x\right)=\sin \left(nx\right)n=1,2,3,\dots$$Now the second case.
$\underset{–}{\lambda =0}$
The general solution is,
$$y\left(x\right)=c_1+c_{2}x$$Applying the first boundary condition gives,
$$\begin{array}{cc}{c}_1+c_2\left(-\pi \right)& =c_1+c_2\left(\pi \right)\\ 2\pi c_2& =0\Rightarrow c_2=0\end{array}$$Using this the general solution is then,
$$y\left(x\right)=c_1$$and note that this will trivially satisfy the second boundary condition just as we saw in the second example above. Therefore, we again have $\lambda =0$ as an eigenvalue for this BVP and the eigenfunctions corresponding to this eigenvalue is,
$$y\left(x\right)=1$$Finally let’s take care of the third case.
$\underset{–}{\lambda <0}$
The general solution here is,
$$y\left(x\right)=c_1\cosh \left(\sqrt{-\lambda }x\right)+c_2\sinh \left(\sqrt{-\lambda }x\right)$$Applying the first boundary condition and using the fact that hyperbolic cosine is even and hyperbolic sine is odd gives,
$$\begin{array}{cc}{c}_1\cosh \left(-\pi \sqrt{-\lambda }\right)+c_2\sinh \left(-\pi \sqrt{-\lambda }\right)& =c_1\cosh \left(\pi \sqrt{-\lambda }\right)+c_2\sinh \left(\pi \sqrt{-\lambda }\right)\\ -c_2\sinh \left(-\pi \sqrt{-\lambda }\right)& =c_2\sinh \left(\pi \sqrt{-\lambda }\right)\\ 2c_2\sinh \left(\pi \sqrt{-\lambda }\right)& =0\end{array}$$Now, in this case we are assuming that $\lambda <0$ and so we know that $\pi \sqrt{-\lambda }\ne 0$ which in turn tells us that $\sinh \left(\pi \sqrt{-\lambda }\right)\ne 0$. We therefore must have $c_2=0$.
Let’s now apply the second boundary condition to get,
$$\begin{array}{cc}\sqrt{-\lambda }{c}_1\sinh \left(-\pi \sqrt{-\lambda }\right)& =\sqrt{-\lambda }{c}_1\sinh \left(\pi \sqrt{-\lambda }\right)\\ 2\sqrt{-\lambda }{c}_1\sinh \left(\pi \sqrt{-\lambda }\right)& =0\Rightarrow c_1=0\end{array}$$By our assumption on $\lambda$ we again have no choice here but to have $c_1=0$.
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.
$$\begin{array}{cccccc}{\lambda }_n& =n^2& y_n\left(x\right)& =\sin \left(nx\right)& n& =1,2,3,\dots \\ {\lambda }_n& =n^2& y_n\left(x\right)& =\cos \left(nx\right)& n& =1,2,3,\dots \\ {\lambda }_0& =0& y_0\left(x\right)& =1& & \end{array}$$Note that we’ve acknowledged that for $\lambda >0$ 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.
$$\begin{array}{cccccc}{\lambda }_n& =n^2& y_n\left(x\right)& =\sin \left(nx\right)& n& =1,2,3,\dots \\ {\lambda }_n& =n^2& y_n\left(x\right)& =\cos \left(nx\right)& n& =0,1,2,3,\dots \end{array}$$

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.