Solving a Sturm-Liouville problem I want to solve the Sturm-Liouville problem:
$\mathcal{A}u=-u''$
with I.C:
\begin{cases}
u(0)=0\\
u'(L)=0
\end{cases}
where $\mathcal{A}$ is the Sturm-Liuville operator defined on $\mathscr{D}_A:\{u\in C^2([0,L])$.
By reading this post from SE , I see that it is practically the same as a PDE. So using the characteristic equation, I obtain:
$u''+Au=0$
$m=\pm\sqrt{-4A}/2=\pm iA$
$u(x)=Be^{iAx}+Ce^{-iAx}$
We rewrite to :
\begin{gather}
u(L)=
\begin{cases}
B\cos Ax\\
iB\sin Ax
\end{cases}
\end{gather}
Since the first conI seedition yields a existence of u at the origin, we can disregard from the cosine solution. We get thus:
\begin{equation}
u(x)=iB\sin Ax
\end{equation}
Using first IC, we have $u(x)=\sin Ax$
Then using the second IC:
\begin{equation}
\begin{array}
fu'(x)=A\cos Ax\\
u'(L)=A\cos A L\\
0=A\cos A x\\
\cos AL =0\\
A=\big(\frac{\pi}{2}+n\pi\big)\frac{1}{L}
\end{array}
\end{equation}
This gives
\begin{equation}
u(x)=\sin \bigg(\big(\frac{\pi}{2}+n\pi\big)\frac{x}{L}\bigg)
\end{equation}
What is wrong?
Thanks!
 A: After you have got $u'(x)=A\cos (Ax)$ you used the second I.C wrong. we know that $u'(L)=0$ so that would give
$$
0=u'(L)=A\cos(AL)\,\Rightarrow\,A=\frac{1}{L}\cdot\left(\frac{\pi}{2}+\pi\cdot n\right)
$$
for $n\in\mathbb{Z}$. It implies that $u(x)=\sin\left(\frac{1}{L}\left(\frac{\pi}{2}+\pi\cdot n\right)x\right)$
A: Your Sturm-Liouville problem is the eigenvalue problem
$$
        -u''(x)=\lambda u(x),\;\;\; 0 \le x \le L
$$
subject to the following endpoint constraints:
$$
      u(0)=0,\;\; u'(L)=0.
$$
The problem is to expand a function $f\in L^2[0,L]$ in terms of the eigenfunction solutions of the above. In your case, you have a regular Sturm-Liouville problem, and the expansion will be a discrete expansion in terms of the eigenfunctions
$$
          e_n(x)=\sin((n+1/2)\pi x/L),\;\; n=1,2,3,\cdots.
$$
It is easy to  verify that $e_n(0)=0$ is satisfied for all $n=1,2,3,\cdots$, and the right endpoint restriction is also satisfied:
$$
   e_n'(L)=\cos((n+1/2)\pi)\{(n+1/2)\pi/L\}=0,\;\; n=1,2,3,\cdots.
$$
The Fourier expansion of $f\in L^2[0,L]$ in terms of these eigenfunctions is given by
$$
     f(x) \sim \sum_{n=1}^{\infty}\frac{\int_{0}^{L}f(y)e_n(y)dy}{\int_{0}^{L}e_n^2(y)dy}e_n(x).
$$
This Sturm-Liouville eigenfunction expansion is guaranteed to converge to $f$ in the norm of $L^2[0,L]$ even though the series expansion always converges to $0$ at $x=0$
