$\partial^2_t u(x,t)=\partial^2_x u(x,t)$ - periodic BC Hi I am looking for a complete solution to the pde given below, it is a hyperbolic pde and I specify the initial conditions and boundary conditions (periodic).  Thanks for your help.
I show what I do below, but I get stuck and must be making a mistake somewhere.  The problem is defined by
\begin{eqnarray}
\partial^2_t u(x,t)=\partial^2_x u(x,t)\\
u(0,t)=u(L,t) \\
\partial_x u(0,t)=\partial_x u(L,t)\\ u(x,0)=\sin(\pi x)\\ \partial_t u(x,0)=0
\end{eqnarray}
where $x\in[0,L], t\geq 0$.  I am looking for an analytical solution, here is my attempt so far (I try separation of variables)
\begin{eqnarray}
u(x,t)=\chi(x)T(t)\implies \frac{T''(t)}{T}=\frac{\chi''(x)}{\chi}=-\alpha^2
\end{eqnarray}
where $-\alpha^2$ is an arbitrary constant, since the function of x and t are equal, they must equal a constant.  Now I obtain
\begin{eqnarray}
T''(t)+\alpha^2 T(t)=0\\
\chi''(x)+\alpha^2 \chi(x)=0
\end{eqnarray}
So I realize I have oscillatory solutions
\begin{eqnarray}
u(x,t)=\chi(x)T(t)=\left[A\cos(\alpha x)+B\sin(\alpha x)\right] \cdot \left[C\cos(\alpha t)+D\sin(\alpha t)\right]
\end{eqnarray}
where $A,B,C,D$ are to be determined from the initial and boundary conditions.
Applying the  initial conditions first, I then have
\begin{eqnarray}
u(x,0)= C [A\cos(\alpha x)+B\sin(\alpha x)]=\sin(\pi x)\\
\implies A=0, \ CB=\alpha =1
\end{eqnarray}
Now imposing the other initial condition I get
\begin{eqnarray}
\partial_t u(x,0)=0= D [B\sin( x)]\implies DB=0
\end{eqnarray}
So I have $A=B=D=0$, so then shouldn't $C=0$ also, which then yields $u(x,t)=0$ which is not right.
I have not yet imposed periodic boundary conditions.  How can I find the solution to this problem?  Thanks!
 A: The solutions $\chi$ should satisfy
$$
     \chi''+\alpha^2\chi = 0\\
     \chi(0)=\chi(L) \\
     \chi'(0)=\chi'(L).
$$
The solutions $\chi(x)=Ae^{i\alpha x}+Be^{-i\alpha x}$ satisfy the periodic conditions iff $\alpha = 2\pi n$ for $n=0,1,2,3,\cdots$. So, $\alpha^2=(2\pi n/L)^2$, and the corresponding solutions are
$$
             \chi_{n}(x) = e^{2\pi inx/L},\;\;\; n=0,\pm 1,\pm 2,\cdots.
$$
The corresponding solutions $T_n(t)$ are $\cos(2n\pi t/L)$ because of the requirement that $T_n'(0)=0$. The general solution is
$$
         u(x,t) = \sum_{n=-\infty}^{\infty}A_n\cos(2\pi nt/L)e^{2\pi inx/L}.
$$
The constants $A_n$ are determined by the initial condition:
$$
     \sin(\pi x)=u(x,0) = \sum_{n=-\infty}^{\infty}A_n e^{2\pi inx/L},\;\;\; 0 < x < L.
$$
Multiplying by $e^{-2\pi inx}$ and integrating over $[0,L]$ gives
$$
       \int_{0}^{L}\sin(\pi x)e^{-2\pi inx/L}dx = A_n L.
$$
The solution:
$$
        u(x,t) = \sum_{n=-\infty}^{\infty}\left(\frac{1}{L}\int_{0}^{L}\sin(\pi y)e^{-2\pi iny/L}dy\right)e^{2\pi inx/L}\cos(2\pi nt/L).
$$
