This question was given in an exam in applied mathematics, on the subject of Fourier series:
Observe the following ODE:
$u\left ( x \right) ^{\prime \prime}+Q \left ( x \right) u\left ( x \right) =f\left ( x \right) $
$u\left ( 0 \right)=u\left ( \pi \right)=0$
Where $Q,f$ are continuous functions which are not periodic. We know that $u \in C^2$ and is not necessarily periodic.
- Let $\frac{a_0}{2} + \sum_{n=1}^{\infty} {a_n \cos{nx}} + \sum_{n=1}^{\infty} {b_n \sin{nx}}$ be $u$'s Fourier series. Show that $a_n=0$ for every $n \geq 0$. That is, $u\left ( x \right) = \sum_{n=1}^{\infty} {b_n \sin{nx}}$.
- Show that $\left | b_n \right | < \frac{c}{n^2}$
What I tried: not a lot. What we are requested to prove in 1 is equivalent to showing that $u$ is an odd function but I cannot see how to we can obtain that from the ODE. It would seem I am missing the right tools to deal with this section and would appreciate it if you could set me on the right path.
$\textbf{Hint:}$ $u$ is not necessary odd, but you can define $2\pi$-periodic odd function $u_1(x)=u(x)$ for $0\leq x \leq \pi$, $u_1=-u(x)$ for $-\pi \leq x \leq 0$. Then you can prove that for $u_1$ you have $a_n=0$, but $u(x)=u_1(x)$ for $0<x<\pi$.