I have to find $\varphi(x)$ such that $$\frac{d^{2}\varphi(x, y)}{dx^{2}}=\gamma\delta(x-y)$$ with the condition $\varphi(x, y)=\varphi(x+L, y)$. The standard approach is to expand in Fourier series $$\varphi(x, y)=\sum_{n\in\mathbb{Z}}\varphi_{n}(y)\exp\Big(\frac{2\pi{i}nx}{L}\Big)$$ This leads to $$-\sum_{n\in\mathbb{Z}, n\neq{0}}\varphi_{n}(y)\Big(\frac{2\pi{n}}{L}\Big)^{2}\exp\Big(\frac{2\pi{i}nx}{L}\Big)=\gamma\delta(x-y)$$ Now we multiply by $\exp\Big(-\frac{2\pi{i}mx}{L}\Big)$ and integrate in $[0, L]$ to give $$-L\varphi_{m}(y)\Big(\frac{2\pi{m}}{L}\Big)^{2}=\gamma\exp\Big(-\frac{2\pi{i}my}{L}\Big)$$ So, the fourier coefficients are $$\varphi_{m}=-\frac{L\gamma}{(2\pi)^{2}}\frac{\exp\Big(-\frac{2\pi{i}my}{L}\Big)}{m^{2}}$$ However, how do we find the $m=0$ term?
2026-03-28 06:48:21.1774680501
Green's function for a 1D poisson equation with periodic boundary condition
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