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I have the Sturm-Liouville problem $$y''(t) + \lambda\ y(t) = 0,\hspace{1cm} y(0) = y(\pi) = 0.$$

When I reach the case where $$\Delta < 0\ \implies \lambda > 0$$ I find $$y(t) = C_1 \cos( \lambda^{\frac{1}{2}} t) + C_2 \sin( \lambda^{\frac{1}{2}} t),$$ and both coefficients $C_1$ and $C_2$ are equal to zero.

How to solve this problem, meaning that I need to find the eigenvalues and eigenfunctions?

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Are you sure that's $y(\pi)=0$, and not $y'(\pi)=0$? –  Ataraxia Jul 27 '13 at 14:27
    
@Ataraxia Yes thats correct –  Geo Papas Jul 27 '13 at 14:28
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@GeoPapas: Can you correct the problem statement if incorrect? You should consider the three cases for $\lambda$. Also, you need to find the eigenfunctions as a conclusion. Regards –  Amzoti Jul 27 '13 at 14:30
    
@Amzoti Yes I have got the other 2 cases but lead to c1=0 and c2=0 and yes, I am trying to find the eigenfunctions. –  Geo Papas Jul 27 '13 at 14:46
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3 Answers

up vote 1 down vote accepted

Since $C_1 = 0$ follows from the first boundary condition, we have $$y(\pi) = 0 = C_2 \sin(\sqrt{\lambda}\pi) \implies \sqrt{\lambda}\pi = n\pi \implies \lambda = n^2 \quad (\text{for } n = 1, 2, \ldots).$$ Now you can find the eigenfunctions because $$y(t) = C_2 \sin(\sqrt{\lambda}t) = C_2 \sin(nt).$$ Note that the case $n = 0$ needs to be studied separately. Just plug in $\lambda = 0$ in the ODE and see what you get. Also, see this related question.

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From $y(0)=0$ you can deduce $C_1=0$. The only option left is $C_2$. See my answer. –  ccorn Jul 27 '13 at 14:43
    
@ccorn: Thanks for the comment. See the edit. –  Mhenni Benghorbal Jul 27 '13 at 14:51
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Note that $n=0$ leads to $y(t)=0$, and zero functions do not count as eigenfunctions (though they are members of eigenspaces, of course). –  ccorn Jul 27 '13 at 15:01
    
@ccorn: In fact, $n=0$ is a special case and it has to be studied separately. –  Mhenni Benghorbal Jul 27 '13 at 15:06
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@MhenniBenghorbal Thanks for the reply! Indeed n=0 leads to y(t)=0 and the eigenfunction is also correct. –  Geo Papas Jul 27 '13 at 18:47
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Your general solution is correct and the given boundary conditions imply $y(t) = 0$. Overall, you have not made any mistakes as you have arrived at the correct solution. However, to solve for the eigenvalues and the eigenfunctions use the fact that $y(\pi) = 0$ and $\sin x = 0 \Longleftrightarrow x = n\pi$ whenever $n \in \mathbb Z$. Do not forget to apply the Superposition Principle. I hope this helps.

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@GeoPapas No mistakes except for the "there is no solution" part. There is a solution, it's $y(t)=0$ :) –  Ataraxia Jul 27 '13 at 14:35
    
@GeoPapas Precisely. –  glebovg Jul 27 '13 at 14:36
    
Those are not inital conditions, but boundary conditions, and these leave possibilities for certain values of $\lambda$. See my answer. –  ccorn Jul 27 '13 at 14:45
    
@ccorn Thanks. By the way, although the general solution will change depending on the sign of $\lambda$, given boundary conditions lead to $y(t) = 0$. –  glebovg Jul 27 '13 at 15:15
    
@glebovg: You essentially state that there are no oscillatory modes for a linearized guitar string. Think again. –  ccorn Jul 27 '13 at 15:15
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With $\lambda=k^2$, $k\in\mathbb{Z}$, $C_2$ may be nonzero. $k=0$ leads to a trivial solution and to $\lambda=0$ which you do not consider anyway. Inferring that you allow arbitrary signs for $\lambda^{\frac{1}{2}}$, let me point out that the solution for $\lambda^{\frac{1}{2}}=+k$ and the one for $\lambda^{\frac{1}{2}} = -k$ are linear dependent, so it suffices to restrict $k$ to be a positive integer.

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Linear dependent after considering the boundary conditions, I mean. –  ccorn Jul 27 '13 at 14:40
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