Differential Equations - Fourier Series: Eigenvalues and Eigenfunctions - iii

Example 4 Find all the eigenvalues and eigenfunctions for the following BVP.$$x^2{y}^{\prime \prime }+3x{y}^{\prime }+\lambda y=0y\left(1\right)=0y\left(2\right)=0$$

This is an Euler differential equation and so we know that we’ll need to find the roots of the following quadratic.
$$r\left(r-1\right)+3r+\lambda =r^2+2r+\lambda =0$$The roots to this quadratic are,
$$r_{1,2}=\frac{-2\pm \sqrt{4-4\lambda }}{2}=-1\pm \sqrt{1-\lambda }$$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 $1-\lambda$. So, let’s go through the cases.
$\underset{–}{1-\lambda <0,\lambda >1}$
In this case the roots will be complex and we’ll need to write them as follows in order to write down the solution.
$$r_{1,2}=-1\pm \sqrt{1-\lambda }=-1\pm \sqrt{-\left(\lambda -1\right)}=-1\pm i\sqrt{\lambda -1}$$By writing the roots in this fashion we know that $\lambda -1>0$ and so $\sqrt{\lambda -1}$ is now a real number, which we need in order to write the following solution,
$$y\left(x\right)=c_1{x}^{-1}\cos \left(\ln \left(x\right)\sqrt{\lambda -1}\right)+c_2{x}^{-1}\sin \left(\ln \left(x\right)\sqrt{\lambda -1}\right)$$Applying the first boundary condition gives us,
$$0=y\left(1\right)=c_1\cos \left(0\right)+c_2\sin \left(0\right)=c_1\Rightarrow c_1=0$$The second boundary condition gives us,
$$0=y\left(2\right)=\frac{1}{2}{c}_2\sin \left(\ln \left(2\right)\sqrt{\lambda -1}\right)$$In order to avoid the trivial solution for this case we’ll require,
$$\sin \left(\ln \left(2\right)\sqrt{\lambda -1}\right)=0\Rightarrow \ln \left(2\right)\sqrt{\lambda -1}=n\pi n=1,2,3,\dots$$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 $\lambda$ gives us the following set of eigenvalues for this case.
$${\lambda }_n=1+{\left(\frac{n\pi }{\ln 2}\right)}^{2}n=1,2,3,\dots$$Note that we need to start the list of $n$’s off at one and not zero to make sure that we have $\lambda >1$ as we’re assuming for this case.
The eigenfunctions that correspond to these eigenvalues are,
$$y_n\left(x\right)=x^{-1}\sin \left(\frac{n\pi }{\ln 2}\ln \left(x\right)\right)n=1,2,3,\dots$$Now the second case.
$\underset{–}{1-\lambda =0,\lambda =1}$
In this case we get a double root of $r_{1,2}=-1$ and so the solution is,
$$y\left(x\right)=c_1{x}^{-1}+c_2{x}^{-1}\ln \left(x\right)$$Applying the first boundary condition gives,
$$0=y\left(1\right)=c_1$$The second boundary condition gives,
$$0=y\left(2\right)=\frac{1}{2}{c}_2\ln \left(2\right)\Rightarrow c_2=0$$We therefore have only the trivial solution for this case and so $\lambda =1$ is not an eigenvalue.
Let’s now take care of the third (and final) case.
$\underset{–}{1-\lambda >0,\lambda <1}$
This case will have two real distinct roots and the solution is,
$$y\left(x\right)=c_1{x}^{-1+\sqrt{1-\lambda }}+c_2{x}^{-1-\sqrt{1-\lambda }}$$Applying the first boundary condition gives,
$$0=y\left(1\right)=c_1+c_2\Rightarrow c_2=-c_1$$Using this our solution becomes,
$$y\left(x\right)=c_1{x}^{-1+\sqrt{1-\lambda }}-c_1{x}^{-1-\sqrt{1-\lambda }}$$Applying the second boundary condition gives,
$$0=y\left(2\right)=c_1{2}^{-1+\sqrt{1-\lambda }}-c_1{2}^{-1-\sqrt{1-\lambda }}=c_1\left(2^{-1+\sqrt{1-\lambda }}-2^{-1-\sqrt{1-\lambda }}\right)$$Now, because we know that $\lambda \ne 1$ 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,
$$c_1=c_2=0$$and so in this case we only have the trivial solution and there are no eigenvalues for which $\lambda <1$.
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 non-trivial 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.$$y^{\prime \prime }+\lambda y=0y\left(0\right)=0y^{\prime }\left(1\right)+y\left(1\right)=0$$

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.
$\underset{–}{\lambda >0}$
The general solution to the differential equation is identical to the first few examples 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,
$$0=y\left(0\right)=c_1\Rightarrow c_1=0$$The second boundary condition gives us,
$$\begin{array}{cc}0=y\left(1\right)+y^{\prime }\left(1\right)& =c_2\sin \left(\sqrt{\lambda }\right)+\sqrt{\lambda }{c}_2\cos \left(\sqrt{\lambda }\right)\\ & =c_2\left(\sin \left(\sqrt{\lambda }\right)+\sqrt{\lambda }\cos \left(\sqrt{\lambda }\right)\right)\end{array}$$So, if we let $c_2=0$ we’ll get the trivial solution and so in order to satisfy this boundary condition we’ll need to require instead that,
$$\begin{array}{cc}0=\sin \left(\sqrt{\lambda }\right)+\sqrt{\lambda }\cos \left(\sqrt{\lambda }\right)& \\ \sin \left(\sqrt{\lambda }\right)& =-\sqrt{\lambda }\cos \left(\sqrt{\lambda }\right)\\ \tan \left(\sqrt{\lambda }\right)& =-\sqrt{\lambda }\end{array}$$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 $\tan \left(\sqrt{\lambda }\right)$ and $-\sqrt{\lambda }$ on the same graph. Here is that graph and note that the horizontal axis really is values of $\sqrt{\lambda }$ 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.
$$\begin{array}{cccccc}\sqrt{{\lambda }_1}& =2.0288& {\lambda }_1& =4.1160& & \left(2.4674\right)\\ \sqrt{{\lambda }_2}& =4.9132& {\lambda }_2& =24.1395& & \left(22.2066\right)\\ \sqrt{{\lambda }_3}& =7.9787& {\lambda }_3& =63.6597& & \left(61.6850\right)\\ \sqrt{{\lambda }_4}& =11.0855& {\lambda }_4& =122.8883& & \left(120.9027\right)\\ \sqrt{{\lambda }_5}& =14.2074& {\lambda }_5& =201.8502& & \left(199.8595\right)\\ \sqrt{{\lambda }_n}& \approx \frac{2n-1}{2}\pi & {\lambda }_n& \approx \frac{{\left(2n-1\right)}^2}{4}{\pi }^2& & n=6,7,8,\dots \end{array}$$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 $n=5$ the error in the approximation is 0.9862%. So less than 1% error by the time we get to $n=5$ and it will only get better for larger value of $n$.
The eigenfunctions for this case are,
$$y_n\left(x\right)=\sin \left(\sqrt{{\lambda }_n}x\right)n=1,2,3,\dots$$where the values of ${\lambda }_n$ are given above.
So, now that all that work is out of the way let’s take a look at 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,
$$0=y\left(0\right)=c_1$$Using this the general solution is then,
$$y\left(x\right)=c_{2}x$$Applying the second boundary condition to this gives,
$$0=y^{\prime }\left(1\right)+y\left(1\right)=c_2+c_2=2c_2\Rightarrow c_2=0$$Therefore, for this case we get only the trivial solution and so $\lambda =0$ is not an eigenvalue. Note however that had the second boundary condition been $y^{\prime }\left(1\right)-y\left(1\right)=0$ then $\lambda =0$ would have been an eigenvalue (with eigenfunctions $y\left(x\right)=x$) 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.
$\underset{–}{\lambda <0}$
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,
$$0=y\left(0\right)=c_1\cosh \left(0\right)+c_2\sinh \left(0\right)=c_1\Rightarrow c_1=0$$Using this the general solution becomes,
$$y\left(x\right)=c_2\sinh \left(\sqrt{-\lambda }x\right)$$Applying the second boundary condition to this gives,
$$\begin{array}{cc}0=y^{\prime }\left(1\right)+y\left(1\right)& =\sqrt{-\lambda }{c}_2\cosh \left(\sqrt{-\lambda }\right)+c_2\sinh \left(\sqrt{-\lambda }\right)\\ & =c_2\left(\sqrt{-\lambda }\cosh \left(\sqrt{-\lambda }\right)+\sinh \left(\sqrt{-\lambda }\right)\right)\end{array}$$Now, by assumption we know that $\lambda <0$ and so $\sqrt{-\lambda }>0$. This in turn tells us that $\sinh \left(\sqrt{-\lambda }\right)>0$ and we know that $\cosh \left(x\right)>0$ for all $x$. Therefore,
$$\sqrt{-\lambda }\cosh \left(\sqrt{-\lambda }\right)+\sinh \left(\sqrt{-\lambda }\right)\ne 0$$and so we must have $c_2=0$ 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 $\lambda >0$ 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.