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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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    This is not the general solution of the equation. It only generates even functions of $x$ (at fixed $t$), whereas the solution can be an arbitrary (twice differentiable) function of $x$ (at fixed $t$).2011-04-04
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    Please enter your equations here so people don't have to click through. You can include $\LaTeX$ between dollar signs.2011-04-04
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    What are your $k_n$? It looks like they should be something like $\frac{2n\pi }{L}$ where $L$ is the length of the box you are in. If you are not in a box you should have in integral instead of a sum.2011-04-04
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    Thank you $K_{n} = \dfrac{n\pi}{L} $2011-04-04
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    .. I'am almost 100% sure that this is the general solution (because it is from a solution written by my teacher)2011-04-04
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    It's not the general solution of the equation alone. There must be some extra conditions given by your teacher that you are not telling us (boundary conditions, initial conditions).2011-04-04
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    I'm sorry for my comments. But the solution is found by using the method: seperation of variables and then using the principle of superposition.2011-04-04
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    The boundery condition is: $\frac{\partial u(x,t)}{\partial x}|_{x=0} = \frac{\partial u(x,t)}{\partial x}|_{x=L} = 0$ And $X\epsilon$ [0,L]2011-04-04
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    If I understand your problem correctly, you are using separation of variables to find a cosine series solution to the wave equation with Neumann initial conditions for $0 \leq x \leq L$. If that is the case, you just need nonnegative $n$, yes.2011-04-04
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    put the boundary conditions into the question2011-04-04
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    Any family of objects indexed by $\mathbb{Z}$, such as your $(C_n,D_n,k_n)$, can be reindexed to yield a family indexed by $\{0,1,2,\ldots\}$ or even by $\{1,2,3,\ldots\}$. Hence, as long as you do not specify $k_n$, the two formulations are equivalent.2011-04-04
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    I suspect that in order to be able to understand any answers given to your question, you will first have to understand why the boundary conditions are an integral part of the definition of the problem (and thus it made no sense to ask the question without mentioning them).2011-04-04
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    @Didier; the $k_n$ had been specified in the meantime (albeit as $K_n$). @Sommerfugl: You should really put all the stuff that you've added in the meantime into the question; you've created quite a mess by not supplying all this essential information right away and then scattering it in comments without fixing the question.2011-04-04
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    @joriki: I know. Es ist egal. (And I agree with your last comment to the OP.)2011-04-04
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    @Sommerfugl: Now you've edited the post to include the boundary conditions, but not the meaning of $k_n$. As Didier rightly pointed out, your question makes no sense if you don't say what $k_n$ is. I think you'll find that people are much more inclined to help you if you make more of an effort to make it easier for them.2011-04-04
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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

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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.