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Suppose I have the following Sturm–Liouville problem

$$\Theta''(\theta) + \lambda \Theta(\theta) = 0$$

with the following conditions

$$\space \Theta(\theta) = \Theta(\theta + 2\pi) $$ $$\space \Theta'(\theta) = \Theta'(\theta + 2\pi)$$ I tried to solve it but I only found that if $\lambda = n^2 \neq 0$ then $\Theta(\theta) = Ae^{in\theta} + Be^{-in\theta}$ with $n \in \mathbb{Z}$ and $\Theta(\theta) = 1$ if $\lambda = 0$. How can I find the values of A and B?

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  • $\begingroup$ They can be anything. $e^{in\theta}$ and $e^{-in\theta}$ are eigenfunctions for this problem. $\endgroup$ – Robert Israel Jan 18 '17 at 16:54
  • $\begingroup$ Why are they each an eigenfunction of this problem and not both together? $\endgroup$ – areobe Jan 18 '17 at 17:03
  • $\begingroup$ Any nonzero linear combination of them is also an eigenfunction. $\endgroup$ – Robert Israel Jan 18 '17 at 17:49

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