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So If I'm looking at a boundary value problem with boundary values of $U_x(0,t) = 0 = U(2\pi,t)$.

I get to $\frac{T'}{kT} = \frac{X"}{x} = \lambda$ and try $\lambda = -\alpha^2$.

This leaves me with $X(x) = A\cos (\alpha x) + B\sin (\alpha x)$ & $X'(x) = -A\alpha\sin (\alpha x) + B\alpha\cos (\alpha x)$

$X(2\pi) = 0 = A\cos(\alpha 2\pi) + B\sin(\alpha 2\pi) = A(1) B B(0) = A$ so $A = 0$.

$X'(0) = 0 = -A\alpha\sin (\alpha 0) + B\alpha\cos (\alpha 0) = A(0) + B(1) = B$ so $B = 0$.

Thus $A = B = 0$, which leads me to believe I have done something wrong here. Any hints or tips you can give me as to what I have done wrong here? I feel confidant I know how to proceed with the problem once I get past this point, however I've been looking at this step for a while now and am not seeing an issue.

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up vote 2 down vote accepted

Your $X'\left(0\right)$ equation correctly gives $B=0$. This leaves the $X\left(2 \pi\right)$ equation, or $X\left(2 \pi \right) = A \cos 2 \pi \alpha = 0$. Now you are supposed to determine the values of $\alpha$ that satisfy this equation. There are infinitely many, and thus infinitely many solutions that you will then sum in a linear combination.

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It's been too long since intro to differential equations I guess. Reading that made a little light go on in my head and sparked my memory, thanks! So then I'd take $\alpha = (2n-1)/4$ for instance and use that to get my eigenvalues for the problem. I guess the A then rolls into my fourier coefficient, right? – Nicholas Hunsicker Nov 13 '12 at 22:33
Right, so you have $\alpha_n = \left(2n-1\right)/4$, and your solution is $U\left(x,t\right) = \sum_n C_n X_n\left(x\right)T_n\left(t\right)$. Now you need a condition on $U\left(x,t_0\right)$ to determine $C_n$. – Eric Angle Nov 14 '12 at 3:22

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