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
$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.