# 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.

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Could you verify the second pair of your BC. I have a strong feeling there ought to be $u_t$ instead of $u_x$. Are you taking the same eigenvalues for both spatial and temporal equation? They must at least have different indexes, say $n$ and $k$. Then you should be able to recover the missing solution from the case $\lambda=0$ – Valentin Aug 3 '12 at 18:30
@Valentin Hi, thank you very much. You are completely right, I made a mistake copying the BC. It's corrected now. About the eigenvalues, in all the exercises I've done, $\lambda_n$ is computed in the spatial equation and then substituted into the temporal one. – Serge Aug 3 '12 at 19:07
@Valentin ...isn't it correct? I have also noticed that, when solving for $T(t)$, the solution assumes $T_0''=0$ so $T_0=A_0t+B_0$. I understand that, but I can't trace the origin of that $T_0''=0$ condition. Also, why should it be added instead of being included in the Fourier Series? Thank you very much. – Serge Aug 3 '12 at 19:16
@Serge I think you are correct - if the eigenvalues for the spatial and temporal equations did not have the same indicies, I don't think the assumption that both sides of the separation of variables equation are equal to the same constant would hold. – Bitrex Aug 3 '12 at 19:34

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.

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thank you very much! As you said, I skipped that $\lambda=0$ solution. I just have a little question concerning the superposition of the solutions. Solving for $\lambda=0$ in both equations, I get that $X(x)=X_0$ and $T(t)=A_0t+B_0$ are solutions. Is it true that $u(x,t)=(A_0t+B_0)X_0$ is a solution for the problem? And then, thanks to linearity we can assume that $X_0=1$. Am I right? So finally we get the most general solution adding the ones for $\lambda>0$. Now that I have written this, it seems completely right. I think I was confusing when I should add or multiply solutions. – Serge Aug 4 '12 at 18:38
Yes you are right. – timur Aug 4 '12 at 18:43
Thank you, timur! – Serge Aug 4 '12 at 19:01

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.

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Thank you very much! – Serge Aug 4 '12 at 18:42