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.