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If I have this equation:

$\frac{\partial^2u}{\partial x^2}=\frac{\partial^2u}{\partial t^2}$

And this general solution:

$u(x,t)=\sum^\infty_{n=-\infty}\cos k_nx(C_n\cos k_nt+D_n\sin k_nt)$

Would it then be wrong to write the above solution with only positive values of $n$? In my text book they often write the result from a superposition with only positive values of $n$, becasue the negative values of n already are included in the terms obtained for positive values of $n$.

The boundary condition:

$\frac{\partial u(x,t)}{\partial x}|_{x=0} = \frac{\partial u(x,t)}{\partial x}|_{x=L} = 0$

To get from $\frac{\partial^2u}{\partial x^2}=\frac{\partial^2u}{\partial t^2}$ to the solution I have used separation of variables and superposition.

Where I found:

U(x,t) = X(x)T(t)

And

X^{''}(x) = -k^{2}X(x)

T^{''}(t) = -k^{2}T(t)

Which gives me:

$X(x) = Acos(kx) + Bsin(kx)$

$T(t) = Ccos(kt) + Dsin(kt)$

Using the boundary condition I get: B = 0 and $k = k_{n} = \dfrac{n\pi}{L}$

Then I use superposition to get the solution:

$u(x,t)=\sum^\infty_{n=-\infty}\cos k_nx(C_n\cos k_nt+D_n\sin k_nt)$

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    But you can find $k_{n}$ from the boundary conditions? But okay I will point it out anyways.2011-04-04

1 Answers 1

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Ignore the chatter, this is a standard undergraduate PDE problem and you're doing it correctly. You may use only $n \geq 0$ as you are doing, and you correctly only have cosine terms in the $x$ variable due to the nature of your boundary conditions.