Green' s function for harmonic oscillator

Question

Does someone know how to get a solution of differential equation for Green's function $(-d^2/dt^2 + \omega^2) G(t, s) = \delta(t-s) $? There is a periodicity of G, actually $\Delta (t-s) = G(t,s)$ and $\Delta (t) = \Delta(t-\beta)$. You should get $\Delta(t)= 1/2\omega [(1+n(\omega)) e^{-\omega t} + n(\omega) e^{\omega t}]$ where $n(\omega)= 1/(e^{\beta \omega} -1)$. Thanks in advance.

Answer 1

You're looking for a $\beta$-periodic function $\Delta$ such that: $\Delta''=\omega^2\Delta$ everywhere except at multiples of $\beta$, $\Delta$ is continuous, and the derivative of $\Delta$ jumps by $-1$ at $0$ (and thus also at all multiples of $\beta$). The RHS is rather $\sum_n\delta(t-s+n\beta)$ (it has to be periodic if the LHS is periodic, and I believe this is what you mean).

So $\Delta$ on $[0,\beta]$ is $\Delta(t)=ae^{\omega t}+be^{-\omega^t}$ for some constants $a,b$, and is extended periodically to all $t$'s. Continuity at $0$ means $a+b=ae^{\omega}+be^{-\omega}$, jump of $\Delta'$ is $(\omega a-\omega b)-(\omega ae^\omega-\omega b e^{-\omega})=-1$. If you solve for $a$ and $b$, you'll get your formula.