I'm hoping to teach myself advanced mathematical methods for physicists and am working my way through Mathematical Analysis of Physical Problems by Philip R. Wallace. I am currently starting on Fourier transforms.
As I read, I work my way through derivations myself with pen and paper because the book sometimes skips quite a few steps for the sake of brevity.
I am currently stuck on a derivation and do not understand where I am going wrong.
Here is the confusing segment:
The Sturm-Liouville problem
$\frac{d^2y}{dx^2}+\lambda y=0$
with the boundary condition of periodicity on the range $(-a/2 \leq x \leq a/2)$:
$y(\frac{a}{2})=y(-\frac{a}{2})$
produces the eigenvalues
$\lambda = \frac{n2\pi}{a}$, $n=integer$
and the corresponding eigenfunctions,
$y=\sin(\frac{n2\pi}{a}x)$
and
$y = \cos(\frac{n2\pi}{a}x)$
which form a degenerate pair. The expansion in these functions
$y=f(x)=\sum_{n=0}^{\infty}(a_n \sin(\frac{n2\pi}{a}x)+b_n \cos(\frac{n2\pi}{a}x))$
is the well-known Fourier series expansion.
Wallace, Philip R., Mathematical analysis of physical problems, New York, NY: Holt, Rinehart & Winston (ISBN 0-03-085626-4). xix, 616 p. (1972). ZBL1092.00500.
I understand the line of thought here, but when I set out to execute this derivation myself, I get a different result for $\lambda$. I first begin by re-arranging the Sturm-Liouville problem:
$\frac{d^2 y}{dx^2} = - \lambda y$
Then I take the approach:
$y(x) = A \sin(\sqrt{\lambda} x) + B \cos(\sqrt{\lambda} x)$.
Now, I use the boundary condition $y(\frac{a}{2})=y(-\frac{a}{2})$ and write
$A \sin(\sqrt{\lambda} \frac{a}{2}) + B \cos(\sqrt{\lambda} \frac{a}{2})=A \sin(-\sqrt{\lambda} \frac{a}{2}) + B \cos(-\sqrt{\lambda} \frac{a}{2})$$
which, considering even/odd symmetry, can be re-written as
$A \sin(\sqrt{\lambda} \frac{a}{2}) + B \cos(\sqrt{\lambda} \frac{a}{2})=-A \sin(\sqrt{\lambda} \frac{a}{2}) + B \cos(\sqrt{\lambda} \frac{a}{2})$.
This gives
$2A \sin(\sqrt{\lambda} \frac{a}{2})=0$.
Assuming the non-trivial case $(A \neq 0)$, I reason that it must be that
$\sqrt{\lambda} \frac{a}{2}=n2\pi$ .
This then yields (assuming only positive $\lambda$):
$\lambda = \frac{n^2 16 \pi^2}{a^2}$.
How is the author coming to his value of $\lambda$? What am I doing wrong?
A Sturm-Liouville problem is a self-adjoint problem. A single condition such as $f(-a/2)=f(a/2)$ is not enough to give you symmetry. For example, define an operator $Lf=-f''$ on the domain $\mathcal{D}(L)$ consisting of twice differentiable functions $f$ on $[-a/2,a/2]$ such that $f(-a/2)=f(a/2)$. A symmetric problem would require that the following would always be $0$ for $f,g\in\mathcal{D}(L)$, and it isn't: \begin{align} \langle Lf,g\rangle-\langle f,Lg\rangle & =\int_{-a/2}^{a/2}-f''\overline{g}+f\overline{g}''dx \\ &= \int_{-a/2}^{a/2}\frac{d}{dx}(-f'\overline{g}+f\overline{g}')dx \\ &= (-f'\overline{g}+f\overline{g}')|_{-a/2}^{a/2}. \end{align} It makes sense that a single condition is not enough because the values of the functions $f,g$ and their first derivatives at $-a/2$ and at $a/2$ come into play in the evaluations on the right. If you impose the additional condition that $f'(-a/2)=f'(a/2)$ for all $f\in\mathcal{D}(L)$, then at least you have $$ \langle Lf,g\rangle = \langle f,Lg\rangle,\;\;\; f,g\in\mathcal{D}(L). $$ That will force eigenvalues of $L$ to be real, and to be discrete, and the eigenfunctions will be a classical Fourier Series invovling $\sin$ and $\cos$ functions. But without the additional condition on $f'$ for $f\in\mathcal{D}(L)$, the problem is not self-adjoint, and you don't get a nice eigenfunction expansion.