On the space of functions $f(x)$ where $x \in (-\pi,\pi]$ satisfying $\int^{\pi}_{-\pi} f(x)dx = 0$ and periodic boundary conditions, find the Green's function for the operator $-\frac{d^2}{dx^2}$.
I know that to solve this problem, homogenous boundary conditions are needed for the operator to be nonsingular, but I am having trouble figuring out how to enforce the boundary conditions. Here is what I do know:
Knowing the properties of Green's function, we can say,
$-\frac{d^2}{dx^2}G(x,\xi) = \delta(x - \xi)$
When $x \neq \xi \rightarrow -\frac{d^2}{dx^2}G(x,\xi) = 0$
From here I am wondering if I would enforce the boundary conditions at both ends of the "crease"; when $x\in (-\pi,\xi]$ and $x\in [\xi,\pi]$. However, I am a little bit unsure what to do with that noninclusive boundary condition.
I have attempted to tackle this problem in a different way by expanding $G(x,\xi)$ as a Fourier series
$G(x,\xi) = \sum_{n = -\infty}^{\infty} g_n e^{in(x-\xi)}$
Then applying the operator:
$-\frac{d^2}{dx^2} \sum_{n = -\infty}^{\infty} g_n e^{in(x-\xi)} = \sum_{n = -\infty}^{\infty} g_n n^2 e^{in(x-\xi)} = \delta(x-\xi)$
Then when compared to the delta function Fourier expansion:
$\delta(x-\xi) = \frac{1}{2\pi}\sum_{n = -\infty}^{\infty} e^{in(x-\xi)}$
We see that $g_n = \frac{1}{2\pi n^2}$. Thus,
$G(x,\xi) = \frac{1}{2\pi} \sum_{n = -\infty}^{\infty} \frac{1}{n^2} e^{in(x-\xi)}$
From here, I don't really know what to do with this infinite series.
So, between these two methods I would really appreciate some guidance on how best to proceed with either method I started this problem with. If something is unclear, please let me know and I will do my best to explain. Thanks in advance!