I am confused by the Sturm-Liouville theorem implication on the Fourier sine and cosine series.
Consider the simple ODE $$y''(x)=-k^2 y(x).$$ It is in Sturm-Liouville form. Let’s impose a boundary condition of $y(0)=0$ and $y(\pi)=0$. The eigen solutions will be $\sin(k x)$.
From the wiki page on the Sturm-Liouville theorem, it is said that “The normalized eigenfunctions form an orthonormal basis under the $w$-weighted inner product in the Hilbert space”.
My first question is: does that mean I can represent any smooth function, in the $(-\pi,\pi)$ region with convergence except at simple discontinuity points?
My second question may seem silly. Consider the function $\sin^2(x)$, it satisfies the boundary conditions above, but it seems cannot be represented by the Fourier sine series. It is obviously represented by a Fourier cosine $1/2-\cos(2x)/2$!
Push that further, if the Fourier sine series is already a complete basis supported by the Sturm-Liouville theorem. How do we represent an even function with a Fourier sine series? The coefficients in tue Fourier sine series evaluated are all zero.
You are discussing conditions at $0$, $\pi$, which means you are looking at convergence in the interval $[0,\pi]$. Sturm-Liouville requires that.
For real $\alpha,\beta$, if you impose general linear endpoint conditions on solutions $f$ of $-f''=\lambda f$ of the form $$ \cos\alpha f(0)+\sin\alpha f'(0) = 0,\\ \cos\beta f(\pi)+\sin\beta f'(\pi)= 0, $$ then the resulting eigenfunctions form an orthogonal basis that can be used to expand a function $f \in L^2[0,\pi]$.