Beam equation and method of separation of variables: some references. Let us consider the beam equation
\begin{align}
z_{tt}+z_{xxxx}&=0\ \ \  \ \ \ \ \ \ 0<t<T, \ 0<x<\pi \notag \\
z(t,0)=z(t,\pi)=z_{xx}(t,0)=z_{xx}(t,\pi)&=0\ \ \  \ \ \ \ \ \ 0<t<T \notag \\
(z(0,.),z_t(0,.))&=(z_0,z_1)
\end{align}
If $z_0=\sum_{k=1}^\infty a_k \sin(kx)$ and $z_1=\sum_{k=1}^\infty b_k \sin(kx)$ the solution of problem has a well-known expression in terms of Fourier series. The method used is, usually, the separation of variables. I'd read in a book or on a website, some notes that explain step by step how you get this solution.
Thank you very much.
 A: Consider $X(x)=a_1\sinh\lambda x+a_2\cosh\lambda x+a_3\sin\lambda x+a_4\cos\lambda x$ ,
$X''(x)=a_1\lambda^2\sinh\lambda x+a_2\lambda^2\cosh\lambda x-a_3\lambda^2\sin\lambda x-a_4\lambda^2\cos\lambda x$
$X(0)=X''(0)=0$ :
$\begin{cases}a_2+a_4=0\\a_2\lambda^2-a_4\lambda^2=0\end{cases}$
$\begin{cases}a_2+a_4=0\\a_2-a_4=0\end{cases}$
$\begin{cases}a_2=0\\a_4=0\end{cases}$
$\therefore X(x)=a_1\sinh\lambda x+a_3\sin\lambda x$
$X(\pi)=X''(\pi)=0$ :
$\begin{cases}a_1\sinh\lambda\pi+a_3\sin\lambda\pi=0\\a_1\lambda^2\sinh\lambda\pi-a_3\lambda^2\sin\lambda\pi=0\end{cases}$
$\begin{cases}a_1\sinh\lambda\pi+a_3\sin\lambda\pi=0\\a_1\sinh\lambda\pi-a_3\sin\lambda\pi=0\end{cases}$
$\begin{cases}a_1\sinh\lambda\pi=0\\a_3\sin\lambda\pi=0\end{cases}$
$\begin{cases}a_1=0\\\sin\lambda\pi=0\end{cases}$
$\begin{cases}a_1=0\\\lambda=n,n\in\mathbb{Z}\end{cases}$
$\therefore$ Let $z(t,x)=\sum\limits_{k=1}^\infty C(t,k)\sin kx$ ,
Then $z_{tt}(t,x)=\sum\limits_{k=1}^\infty C_{tt}(t,k)\sin kx$
$z_{xxxx}(t,x)=\sum\limits_{k=1}^\infty k^4C(t,k)\sin kx$
$\therefore\sum\limits_{k=1}^\infty C_{tt}(t,k)\sin kx+\sum\limits_{k=1}^\infty k^4C(t,k)\sin kx=0$
$\sum\limits_{k=1}^\infty(C_{tt}(t,k)+k^4C(t,k))\sin kx=0$
$\therefore C_{tt}(t,k)+k^4C(t,k)=0$
$C(t,k)=A(k)\sin k^2t+B(k)\cos k^2t$
$\therefore z(t,x)=\sum\limits_{k=1}^\infty A(k)\sin k^2t\sin kx+\sum\limits_{k=1}^\infty B(k)\cos k^2t\sin kx$
$z_t(t,x)=\sum\limits_{k=1}^\infty A(k)k^2\cos k^2t\sin kx-\sum\limits_{k=1}^\infty B(k)k^2\sin k^2t\sin kx$
$z(0,x)=z_0$ , $z_t(0,x)=z_1$ :
$\begin{cases}\sum\limits_{k=1}^\infty A(k)k^2\sin kx=z_1\\\sum\limits_{k=1}^\infty B(k)\sin kx=z_0\end{cases}$
$\begin{cases}A(k)=\dfrac{2}{k^2\pi}\int_0^\pi z_1\sin kx~dx\\B(k)=\dfrac{2}{\pi}\int_0^\pi z_0\sin kx~dx\end{cases}$
$\therefore z(t,x)=\dfrac{2}{\pi}\sum\limits_{k=1}^\infty\dfrac{1}{k^2}\int_0^\pi z_1\sin kx~dx\sin k^2t\sin kx+\dfrac{2}{\pi}\sum\limits_{k=1}^\infty\int_0^\pi z_0\sin kx~dx\cos k^2t\sin kx$
