For basic Fourier series theory we will need the following three eigenvalue problems. We will consider more general equations and boundary conditions, but we will postpone this until
Chapter 5.
\begin{equation}
x'' + \lambda x = 0, \quad x(a) = 0, \quad x(b) = 0 ,\tag{4.1}
\end{equation}
\begin{equation}
x'' + \lambda x = 0, \quad x'(a) = 0, \quad x'(b) = 0 ,\tag{4.2}
\end{equation}
and
\begin{equation}
x'' + \lambda x = 0, \quad x(a) = x(b), \quad x'(a) = x'(b) .\tag{4.3}
\end{equation}
A number
\(\lambda\) is called an
eigenvalue of
(4.1) (resp.
(4.2) or
(4.3)) if and only if there exists a nonzero (not identically zero) solution to
(4.1) (resp.
(4.2) or
(4.3)) given that specific
\(\lambda\text{.}\) A nonzero solution is called a corresponding
eigenfunction.
Note the similarity to eigenvalues and eigenvectors of matrices. The similarity is not just coincidental. If we think of the equations as differential operators, then we are doing the same exact thing. Think of a function \(x(t)\) as a vector with infinitely many components (one for each \(t\)). Let \(L = -\frac{d^2}{{dt}^2}\) be the linear operator. Then the eigenvalue/eigenfunction pair should be \(\lambda\) and nonzero \(x\) such that \(Lx = \lambda x\text{.}\) In other words, we are looking for nonzero functions \(x\) satisfying certain endpoint conditions that solve \((L- \lambda)x = 0\text{.}\) A lot of the formalism from linear algebra still applies here, though we will not pursue this line of reasoning too far.
Example 4.1.3.
Let us find the eigenvalues and eigenfunctions of
\begin{equation*}
x'' + \lambda x = 0, \quad x(0) = 0, \quad x(\pi) = 0 .
\end{equation*}
We have to handle the cases \(\lambda > 0\text{,}\) \(\lambda = 0\text{,}\) \(\lambda < 0\) separately. First suppose that \(\lambda > 0\text{.}\) Then the general solution to \(x''+\lambda x = 0\) is
\begin{equation*}
x = A \cos \bigl( \sqrt{\lambda}\, t\bigr)
+ B \sin \bigl( \sqrt{\lambda}\, t\bigr).
\end{equation*}
The condition \(x(0) = 0\) implies immediately \(A = 0\text{.}\) Next
\begin{equation*}
0 = x(\pi) = B \sin \bigl( \sqrt{\lambda}\, \pi \bigr) .
\end{equation*}
If \(B\) is zero, then \(x\) is not a nonzero solution. So to get a nonzero solution we must have that \(\sin \bigl( \sqrt{\lambda}\, \pi\bigr) = 0\text{.}\) Hence, \(\sqrt{\lambda}\, \pi\) must be an integer multiple of \(\pi\text{.}\) In other words, \(\sqrt{\lambda} = k\) for a positive integer \(k\text{.}\) Hence the positive eigenvalues are \(k^2\) for all integers \(k \geq 1\text{.}\) Corresponding eigenfunctions can be taken as \(x=\sin (k t)\text{.}\) Just like for eigenvectors, constant multiples of an eigenfunction are also eigenfunctions, so we only need to pick one.
Now suppose that \(\lambda = 0\text{.}\) In this case the equation is \(x'' = 0\text{,}\) and its general solution is \(x = At + B\text{.}\) The condition \(x(0) = 0\) implies that \(B=0\text{,}\) and \(x(\pi) = 0\) implies that \(A = 0\text{.}\) This means that \(\lambda
= 0\) is not an eigenvalue.
Finally, suppose that \(\lambda < 0\text{.}\) In this case we have the general solution
\begin{equation*}
x = A \cosh \bigl( \sqrt{-\lambda}\, t\bigr)
+ B \sinh \bigl( \sqrt{-\lambda}\, t \bigr) .
\end{equation*}
Letting \(x(0) = 0\) implies that \(A = 0\) (recall \(\cosh 0 = 1\) and \(\sinh 0 =
0\)). So our solution must be \(x = B \sinh \bigl( \sqrt{-\lambda}\, t \bigr)\) and satisfy \(x(\pi) = 0\text{.}\) This is only possible if \(B\) is zero. Why? Because \(\sinh \xi\) is only zero when \(\xi=0\text{.}\) You should plot sinh to see this fact. We can also see this from the definition of sinh. We get \(0 = \sinh \xi = \frac{e^\xi -
e^{-\xi}}{2}\text{.}\) Hence \(e^\xi = e^{-\xi}\text{,}\) which implies \(\xi = -\xi\) and that is only true if \(\xi=0\text{.}\) So there are no negative eigenvalues.
In summary, the eigenvalues and corresponding eigenfunctions are
\begin{equation*}
\lambda_k = k^2 \qquad \text{with an eigenfunction} \qquad x_k = \sin (k t)
\qquad \text{for all integers } k \geq 1 .
\end{equation*}
Example 4.1.4.
Let us compute the eigenvalues and eigenfunctions of
\begin{equation*}
x'' + \lambda x = 0, \quad x'(0) = 0, \quad x'(\pi) = 0 .
\end{equation*}
Again we have to handle the cases \(\lambda > 0\text{,}\) \(\lambda = 0\text{,}\) \(\lambda
< 0\) separately. First suppose that \(\lambda > 0\text{.}\) The general solution to \(x''+\lambda x = 0\) is \(x = A \cos \bigl( \sqrt{\lambda}\, t\bigr) + B \sin \bigl( \sqrt{\lambda}\,
t\bigr)\text{.}\) So
\begin{equation*}
x' = -A\sqrt{\lambda}\, \sin \bigl( \sqrt{\lambda}\, t\bigr) +
B\sqrt{\lambda}\, \cos \bigl(\sqrt{\lambda}\, t\bigr) .
\end{equation*}
The condition \(x'(0) = 0\) implies immediately \(B = 0\text{.}\) Next
\begin{equation*}
0 = x'(\pi) = -A\sqrt{\lambda}\, \sin \bigl( \sqrt{\lambda}\, \pi\bigr) .
\end{equation*}
Again \(A\) cannot be zero if \(\lambda\) is to be an eigenvalue, and \(\sin \bigl( \sqrt{\lambda}\, \pi\bigr)\) is only zero if \(\sqrt{\lambda} = k\) for a positive integer \(k\text{.}\) Hence the positive eigenvalues are again \(k^2\) for all integers \(k \geq 1\text{.}\) And the corresponding eigenfunctions can be taken as \(x=\cos (k t)\text{.}\)
Now suppose that \(\lambda = 0\text{.}\) In this case the equation is \(x'' = 0\) and the general solution is \(x = At + B\) so \(x' = A\text{.}\) The condition \(x'(0) = 0\) implies that \(A=0\text{.}\) The condition \(x'(\pi) = 0\) also implies \(A=0\text{.}\) Hence \(B\) could be anything (let us take it to be 1). So \(\lambda = 0\) is an eigenvalue and \(x=1\) is a corresponding eigenfunction.
Finally, let \(\lambda < 0\text{.}\) In this case the general solution is \(x = A \cosh \bigl( \sqrt{-\lambda}\, t\bigr)
+ B \sinh \bigl( \sqrt{-\lambda}\, t\bigr)\) and
\begin{equation*}
x' = A\sqrt{-\lambda}\, \sinh \bigl( \sqrt{-\lambda}\, t\bigr)
+ B\sqrt{-\lambda}\, \cosh \bigl( \sqrt{-\lambda}\, t \bigr) .
\end{equation*}
We have already seen (with roles of \(A\) and \(B\) switched) that for this expression to be zero at \(t=0\) and \(t=\pi\text{,}\) we must have \(A=B=0\text{.}\) Hence there are no negative eigenvalues.
In summary, the eigenvalues and corresponding eigenfunctions are
\begin{equation*}
\lambda_k = k^2 \qquad \text{with an eigenfunction} \qquad x_k = \cos (k t)
\qquad \text{for all integers } k \geq 1 ,
\end{equation*}
and there is another eigenvalue
\begin{equation*}
\lambda_0 = 0 \qquad \text{with an eigenfunction} \qquad x_0 = 1.
\end{equation*}
The following problem is the one that leads to the general Fourier series.
Example 4.1.5.
Let us compute the eigenvalues and eigenfunctions of
\begin{equation*}
x'' + \lambda x = 0, \quad x(-\pi) = x(\pi), \quad x'(-\pi) = x'(\pi) .
\end{equation*}
We have not specified the values or the derivatives at the endpoints, but rather that they are the same at the beginning and at the end of the interval.
Let us skip \(\lambda < 0\text{.}\) The computations are the same as before, and again we find that there are no negative eigenvalues.
For \(\lambda = 0\text{,}\) the general solution is \(x = At + B\text{.}\) The condition \(x(-\pi) = x(\pi)\) implies that \(A=0\) (\(A\pi + B = -A\pi +B\) implies \(A=0\)). The second condition \(x'(-\pi) = x'(\pi)\) says nothing about \(B\) and hence \(\lambda=0\) is an eigenvalue with a corresponding eigenfunction \(x=1\text{.}\)
For \(\lambda > 0\) we get that \(x = A \cos \bigl( \sqrt{\lambda}\, t \bigr)
+ B \sin \bigl( \sqrt{\lambda}\, t\bigr)\text{.}\) Now
\begin{equation*}
\underbrace{A \cos \bigl(-\sqrt{\lambda}\, \pi\bigr)
+ B \sin \bigl(-\sqrt{\lambda}\,
\pi\bigr)}_{x(-\pi)}
=
\underbrace{A \cos \bigl( \sqrt{\lambda}\, \pi \bigr)
+ B \sin \bigl( \sqrt{\lambda}\,
\pi\bigr)}_{x(\pi)} .
\end{equation*}
We remember that \(\cos (- \theta) = \cos (\theta)\) and \(\sin (-\theta) = - \sin (\theta)\text{.}\) Therefore,
\begin{equation*}
A \cos \bigl(\sqrt{\lambda}\, \pi\bigr)
- B \sin \bigl( \sqrt{\lambda}\, \pi\bigr)
=
A \cos \bigl(\sqrt{\lambda}\, \pi\bigr)
+ B \sin \bigl( \sqrt{\lambda}\, \pi\bigr).
\end{equation*}
Hence either \(B=0\) or \(\sin \bigl( \sqrt{\lambda}\, \pi\bigr) = 0\text{.}\) Similarly (exercise) if we differentiate \(x\) and plug in the second condition we find that \(A=0\) or \(\sin \bigl( \sqrt{\lambda}\, \pi\bigr) = 0\text{.}\) Therefore, unless we want \(A\) and \(B\) to both be zero (which we do not) we must have \(\sin \bigl( \sqrt{\lambda}\, \pi \bigr) = 0\text{.}\) Hence, \(\sqrt{\lambda}\) is an integer and the eigenvalues are yet again \(\lambda = k^2\) for an integer \(k \geq 1\text{.}\) In this case, however, \(x = A \cos (k t) + B \sin (k t)\) is an eigenfunction for any \(A\) and any \(B\text{.}\) So we have two linearly independent eigenfunctions \(\sin (kt)\) and \(\cos (kt)\text{.}\) Remember that for a matrix we can also have two eigenvectors corresponding to a single eigenvalue if the eigenvalue is repeated.
In summary, the eigenvalues and corresponding eigenfunctions are
\begin{equation*}
\begin{aligned}
& \lambda_k = k^2 & & \text{with eigenfunctions} & &
\cos (k t) \quad \text{and}\quad \sin (k t)
& & \text{for all integers } k \geq 1 , \\
& \lambda_0 = 0 & & \text{with an eigenfunction} & & x_0 = 1.
\end{aligned}
\end{equation*}
