Wave Equation Neumann Boundary Conditions I am studying basic PDEs and I would like to ask a thing I can't understand. I would really appreciate a piece of advice. I must compute the solution $u(x,t)$ of a 1-D wave equation with Neumann boundary conditions:
$u_{tt}= u_{xx} $ , $0<x<1 $
$u_{x}(0,t) = u_{x}(1,t) = 0 $
$u(x,0) = x $
$u_{t}(x,0) = 1$
Separating variables, I get to two independent differential equations.
$ X''(x) = - \lambda X(x)$
$ T''(x) = - \lambda T(t)$
First, I solve for the spatial one, getting the eigenvalues $\lambda_n = (n\pi)^2$ and eigenfunctions $\varphi_n=\cos(n\pi x)$.
Then, substituting $\lambda_n$, the family of solutions for the temporal equation shall be $T_n(t)=C_n \cos(n\pi t) + D_n \sin(n\pi t)  $
I think everything is OK up to this point. I would apply superposition to get $u(x,t) = \sum_{n=1}^\infty (C_n \cos(n\pi t) + D_n \sin(n\pi t))cos(n\pi x)$
but, and here is my question, the solution applies the BC before that:
$u_{x}(x,0) = 1 $ so $u_{xx}(x,0) = 0 $,
thus $T_0=A_0t+B_0 $
and therefore, reaching a different solution,
$u(x,t) = A_0t+B_0  +\sum_{n=1}^\infty (C_n \cos(n\pi t) + D_n \sin(n\pi t))\cos(n\pi x)$
And finally, the solution computes that $B_0$ as the first order cosine series coefficient, and gets $A_0=1$ from the $u_{x}(x,0) = 1$ condition.
I don't know why should that steps be done. I don't understand why do we need to add that $T_0$. In fact, why do we want to sum it to the solution?
On the other hand, this TTU paper doesn't use that $T_0$.
Thank you very much for your time!
PS: Please feel free to tell me if you find any kind of inconsistency. I have not been feeling very confident about my English lately.
 A: When you separate variables, you are led to the equations 
$ X''(x) = - \lambda X(x)$
$ T''(x) = - \lambda T(t)$
to be solved with some constant $\lambda$. Note that there is no restriction on the sign of $\lambda$. Any restriction must come from the problem itself, rather than being arbitrarily imposed by us. Then you solve the spatial part, and realize that there is no nontrivial solution if $\lambda<0$. The case $\lambda=0$ gives the solution $X(x)=1$, which leads to $T_0$. The last remaining case $\lambda>0$ implies the condition that $\lambda$ must be of the form $(n\pi)^2$, where $n$ is a positive integer. What you are doing in the last step is simply collecting all the solutions you obtained.
The TTU paper you linked to is somewhat incomplete. Perhaps you have to read that in light of the earlier notes in the series.
A: Since it is not possible to hear the shape of a drum it is helpful to specify which domain you are given. I shall take it this is rectangular in your case. First of all it is useful to write out the boundary conditions for the "separated" problems.
For instance, the first pair implies
$$X'(0)=X'(1)=0$$
The very fact that
$$\frac{X''}{X}=-\lambda=\text{const}\tag{1}$$
does not say anything about the sign of $\lambda$, so you have to consider separately the cases where $\lambda<0$, $\lambda=0$, $\lambda>0$ and see which of them fit the boundary conditions. For example, if $\lambda<0$ then $(1)$ gives:
$$X=Ae^x+Be^{-x}$$
Applying boundary conditions we obtain $A=B=0$, hence there are no nontrivial solutions. Therefore $\lambda=\omega^2$
$$X=A\cos\omega x+B\sin\omega x$$
$$X'=-A\omega\sin\omega x+B\omega\cos\omega x$$
Applying boundary conditions we find
$$\sin\omega x=0$$
$$\omega=\pi n,\quad n\in\mathbb{Z}$$
$$\lambda = \pi^2n^2$$
The key idea behind separating variables is the assumption that the two parts are independent. Therefore, you must choose an independent index to enumerate eigenvalues, otherwise you will not get all solutions. If you carry out similar calculations carefully for the $T$ part you should get the result agreeing with the given solution.
