Periodic solutions to the wav equation with seperated solutions Consider the wave equation $ u_{tt}=a^2 u_{xx} $ and a separated solution
$u(x,t)=T(t) \varphi (x) $, with boundary condition $u(0,t)=c, \ u(1,t)=d$.
Then I want to show that all separated solutions are periodic in both $x$ and $t$.
From lectures I know its true for $c=d=0$ so I can assume one is non zero. Thus assume $c \neq 0$. 
By plugging in the separated solution to our pde we get
$- \frac{ \varphi''(x)}{ \varphi (x)}=-\frac{T''(t)}{a^2 T(t)}=\lambda $
This gives two associated odes
$\varphi''(x)+\lambda \varphi(x)=0 \quad  T''(t)+\lambda a^2 T(t)=0   $
one can easily show that $\lambda >0$ thus set $\lambda = \beta^2$. One can also easily show that our two solutions are
$\varphi(x)=A \cos(\beta x)+B \sin(\beta x), \quad T(t)=C \cos(\beta a t)+D \sin(\beta at),$
My problem is that my boundary conditions don't give me much to work with, they give the following:
$u(0,t)=A(t) \varphi (0)=A \cdot T(t)=c \ $ and $ \ u(1,t)=\varphi (1) T(t)=d$
I cant really use these to much, and I get the hint that I should see that $T(t)$ is a periodic function or $T(t) \rightarrow \pm \infty$ for $t \rightarrow \infty$. I believe my problems lies in understanding the boundaries conditions, or maybe I have made some mistake?
 A: For the pde 
\begin{align}
u_{tt} = a^{2} u_{xx} \hspace{10mm} u(0,t) = c, \hspace{5mm} u(1,t) = d
\end{align}
consider the separation function $u(x,t) = T(t) \phi(x)$ can be used to obtain the differential equations 
\begin{align}
\phi(x) &= A_{0} \cos(\beta x) + B_{0} \sin(\beta x) \\
T(t) &= C_{0} \cos(\beta a t) + D_{0} \sin(\beta at).
\end{align}
From the boundary conditions of $u(x,t)$ it is given that $\phi(0) = c$ and $\phi(1) = d$. This leads to
\begin{align}
\phi(0) = c &= A_{0} \\
\phi(1) = d &= A_{0} \cos(\beta) + B_{0} \sin(\beta).
\end{align}
This process "seems" to lead to much more confusion. So, try a different method. 
Consider changing the boundary conditions by a function that has a value of zero for a second order derivative. In this view consider the form
\begin{align}
u(x,t) = c + (d-c)x + \phi(x) \psi(t).
\end{align}
Notice that the values of the boundary conditions are built in to the solution. This can be checked as a valid solution. What is changed is that now it can be stated that the conditions become
\begin{align}
\phi(0) = 0  \hspace{10mm} \phi(1) = 0.
\end{align}
Now the $x$ differential equation can be seen to follow
\begin{align}
\phi(0) = 0 &= A_{0} \\
\phi(1) = 0 &= B_{0} \sin(\beta).
\end{align}
Either $B_{0} = 0$, which is possible, or $\sin(\beta) = 0$ yielding $\beta = n \pi$. With this the solution then becomes
\begin{align}
u(x,t) = c + (d-c) x + \sum_{n=1}^{\infty} \sin(n \pi x) \left( A_{n} \cos(n \pi t) + B_{n} \sin(n \pi t) \right). 
\end{align}
In order to further reduce the general solution some condition for $t$ is needed.
