Eigenvalue problem with periodic boundary conditions-complete Fourier series
$$y''+\lambda y=0, 0 \leq x \leq L$$ $$(*): \begin{cases} y(0)=y(L)\\[4pt] y'(0)=y'(L) \end{cases}$$ $$$$ It's a Sturm-Liouville problem.
$$y(x)=c \sin (\sqrt{\lambda }x)+d \cos( \sqrt{\lambda }x)$$ $$y'(x)=\sqrt{\lambda} c \cos(\sqrt{\lambda} x)-d \sqrt{\lambda} \sin( \sqrt{\lambda}x)$$
$(*):$$$y(0)=y(L) \Rightarrow d =c \sin (\sqrt{\lambda}L)+d \cos(\sqrt{\lambda}L)$$ $$y'(0)=y'(L) \Rightarrow c =c \cos(\sqrt{\lambda}L)-d \sin(\sqrt{\lambda}L)$$
Why do we want that the determinant of the coefficients is $0$? $$\det\begin{bmatrix} \sin(\sqrt{\lambda}L) & \cos(\sqrt{\lambda}L)-1\\ \cos(\sqrt{\lambda}L)-1 & -\sin(\sqrt{\lambda}L) \end{bmatrix}=0$$
$$$$
EDIT:
I tried to find the eigenvalues/eigenfunctions in an other way...
Could you tell me if it's also correct??
At the case $\lambda >0: \\ $ $$y(x)=c_1 \cos{(\sqrt{\lambda} x)}+c_2 \sin{(\sqrt{\lambda}x)}$$ $$y(0)=y(L) \Rightarrow c_1=c_1 \cos{(\sqrt{\lambda}L)}+c_2 \sin{(\sqrt{\lambda} L)} \ (1)$$ $$y'(0)=y'(L) \Rightarrow c_2=-c_1 \sin{(\sqrt{\lambda}L)}+c_2 \cos{(\sqrt{\lambda}L)} \ (2)$$ $$c_2(1)-c_1(2): 0=c_2^2 \sin{(\sqrt{\lambda} L)}+c_1^2 \sin{(\sqrt{\lambda} L)} \Rightarrow \sin{(\sqrt{\lambda}L)} (c_1^2+c_2^2)=0 \Rightarrow \sin{(\sqrt{\lambda}L)}=0 \Rightarrow \sqrt{\lambda}=\frac{n \pi}{L}$$
Is this correct?? In my notes, with the first way, it is $$\sqrt{\lambda}=\frac{2 n \pi}{L}$$ Which of them is correct??
The first way is correct. Assume that the second way is correct too. If we substitute $$ \sqrt{\lambda} = \frac{\pi}{L} $$ to the boundary condition $y(0) = y(L)$, we get $$ c_1 = -c_1. $$ This is possible only for $c_1 =0$. However, in this case the second boundary condition will get us $c_2 = -c_2$, which imply $c_2 = 0$. Thus, we get $(c_1, c_2) = (0, 0)$, which is non-interesting case.