Periodic boundary conditions

Remarks

Example: on $[0, 2\pi]$, the function $y(x) = \sin x$ satisfies $y(0) = 0 = y(2\pi)$ and $y'(0) = 1 = y'(2\pi)$. So it satisfies the periodic boundary conditions, and it solves $y'' + y = 0$. Non-example: on $[0, \pi]$, the same function has $y(0) = y(\pi) = 0$ but $y'(0) = 1 \ne -1 = y'(\pi)$, so the second condition fails. The eigenvalue problem $-y'' = \lambda y$ with periodic boundary conditions on $[0, 2\pi]$ has the eigenvalues $\lambda = k^2$ for $k = 0, 1, 2, \ldots$. The eigenvalue $0$ has the eigenfunction $1$, and each $k^2$ with $k \ge 1$ has the two independent eigenfunctions $\cos kx$ and $\sin kx$. Unlike separated conditions, periodic conditions couple the two endpoints, so eigenvalues of multiplicity $2$ can occur. If the coefficients of the equation are $(b - a)$-periodic, a solution with these conditions extends to a $(b - a)$-periodic solution on $\R$.