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Find the eigenfunctions and eigenvalues of $y" + \lambda y = 0$ with $y(L) = y(0)$ and $y'(L) = y'(0)$.

I broke the problem into three parts, depending on the sign of $\lambda$ (although $\lambda = 0$ and $\lambda < 0$ turned out to be the same). For $\lambda \le 0$, I found that the eigenfunctions were $y_n = C$ where $C$ is a constant, with associated eigenvalue $\lambda_n = 0$.

I ran into difficulty in the case of $\lambda > 0$. Here, I let $\lambda = \mu^2$. $$y = A\sin(\mu x) + B\cos(\mu x)$$ The initial conditions yield: $$A\sin(\mu L) + B\cos(\mu L) = B$$ $$A\cos(\mu L) - B\sin(\mu L) = A$$ Then $$A\sin(\mu L) + B\cos(\mu L) - B = A\cos(\mu L) - B\sin(\mu L) - A$$ $$A(1 + \sin(\mu L) - \cos(\mu L)) = B(1 - cos(\mu L) - sin(\mu L))$$ Therefore, the eigenfunctions are $$y_n = B\left(\frac{1 - \cos(\mu_n L) - \sin(\mu_n L)}{1 + \sin(\mu_n L) - \cos(\mu_n L)}\sin(\mu_n x) + \cos(\mu_n x)\right)$$ where $\mu_n$ can (apparently) be any positive real number (the eigenvalues will be $\sqrt{\mu}$.

I'm don't believe this is correct. I've checked over my work many times, and don't see an error. However, it seems strange to me that the eigenfunctions have a continuous spectra. I've never encountered that before in a Sturm-Liouville problem. Moreover, it doesn't appear to satisfy the initial conditions. Any help would be greatly appreciated!

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    $\mu$ can't be arbitrary since your solutions should have period $L$ – 2017-01-31
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    @user399601: Yes, I understand that part. I don't see how to get any other conclusion from the work I've done. Could you perhaps tell me *why* it's wrong? – 2017-01-31
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    If $B \ne 0$ then the only solutions to $A \sin(\mu L) + B \cos(\mu L) = B$ and $A \cos(\mu L) - B \sin(\mu L) = A$ are when $\sin(\mu L) = 0$ and $\cos(\mu L) = 1.$ The rest of your work involves dividing by $0$. – 2017-01-31

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Your initial conditions gave $$ A\sin(\mu L)+B\cos(\mu L)= B \\ A\cos(\mu L)-B\sin(\mu L)= A. $$ or $$ \left[\begin{array}{cc}\sin(\mu L) & (\cos(\mu L)-1) \\ (\cos(\mu L)-1) & -\sin(\mu L)\end{array}\right] \left[\begin{array}{c} A \\ B \end{array}\right] = \left[\begin{array}{c}0 \\ 0 \end{array}\right]. $$ There is a non-trivial solution iff the determinant $D(\lambda)$ of the coefficient matrix is $0$, which is $$ D(\lambda) = -\sin^2(\mu L)-(\cos(\mu L)-1)^2 = 2(\cos(\mu L)-1). $$ These solutions are $\mu_n = \frac{1}{L}2n\pi$, which leads to the $0$ coefficient matrix, meaning that $A$ and $B$ are arbitrary in this case. In other words $A\cos(\mu_n x)+B\sin(\mu_n x)$ is an eigenfunction solution for arbitrary $A$, $B$.

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    If $\mu_n = \frac{1}{L}(\frac{\pi}{2} + n \pi)$, then $\cos(\mu L) = 0$, so that $D(\lambda) = -2$. Perhaps you meant $\mu_n = \frac{2 \pi}{L}n$ ? – 2017-02-01
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    @infinitylord : Yes, you're right. – 2017-02-01